Given: R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 4 Ω, R₄ = 6 Ω, R₅ = 10 Ω, V = 20 V

Step 1: R₃ and R₄ are in series → R₃₄ = 4 + 6 = 10 Ω

Step 2: R₂ (10 Ω) and R₃₄ (10 Ω) are in parallel:
$$\frac{1}{R_{parallel}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \Rightarrow R_{parallel} = 5 \text{ Ω}$$

(i) Effective resistance:
R₁ is in series with R_parallel:
$$R_{eff} = 5 + 5 = \boxed{10 \text{ Ω}}$$

(ii) Current drawn from battery (Ohm's law):
$$I = \frac{V}{R_{eff}} = \frac{20}{10} = \boxed{2 \text{ A}}$$

(iii) Potential difference across R₁ (5 Ω):
$$V_1 = I \times R_1 = 2 \times 5 = \boxed{10 \text{ V}}$$

Source: Chapter 11, Sections 11.6.1 and 11.6.2

---

• Circuit reading: Identify series/parallel groupings carefully. R₃ + R₄ are in series (same branch), their combination is parallel to R₂, and the whole parallel block is in series with R₁.
• Examiner looks for: correct identification of combination type, proper formula application, and 3 clearly labelled answers — one for each sub-part.
• Common mistake: Adding all resistors directly or misidentifying which resistors are parallel. Always reduce step by step.
• The voltmeter reads V across R₁ only (it is connected across R₁), so the answer to (iii) is simply I × R₁.

Given: P = 1100 W, V = 220 V, time = 5 h/day for 6 days

(i) Current drawn from mains:

Using P = VI,

$$I = \frac{P}{V} = \frac{1100 \text{ W}}{220 \text{ V}} = 5 \text{ A}$$

(ii) Energy consumed:

Total time = 5 × 6 = 30 hours

$$\text{Energy} = P \times t = 1100 \text{ W} \times 30 \text{ h} = 33000 \text{ Wh} = 33 \text{ kWh}$$

(iii) Total cost:

$$\text{Cost} = 33 \text{ units} \times ₹5 = ₹165$$

Source: Chapter 11, Section 11.8 Electric Power

---

• Examiners expect the formula to be stated, values substituted, and units written clearly for each part.
• 1 kWh = 1 unit; always convert Wh → kWh before calculating cost.
• Budget: ~1 mark per sub-part — keep each part to 2–3 lines.

(a) These gadgets should be connected in parallel with the mains. Three reasons:

1. Each appliance gets the same voltage (220 V) as required for proper functioning.
2. If one appliance is switched off, others continue to work independently.
3. The total resistance decreases in parallel, allowing each device to draw the required current without affecting others.

(b) An electric fuse is a safety device made of a thin wire of low melting point alloy (e.g., tin-lead) that melts and breaks the circuit when excess current flows, preventing damage to appliances.
It is always connected in series with the live wire of the circuit.

---

• For part (a), the three reasons — same voltage, independent operation, and individual switching — are standard CBSE expected points. Any three valid reasons score full marks.
• For part (b), examiners look for: definition mentioning low melting point, its protective function, and that it is connected in series. Missing "series" is a common mistake and will cost marks.
• The question is 3 marks total, so keep answers brief and to the point.

(a) Electric power of an electrical device is the rate at which it consumes electrical energy.
$$P = \frac{W}{t} = VI$$
SI unit: Watt (W).

(b) Given: Power P = 2 kW, Time t = 2 h

(i) Energy in kilowatt hour:
$$E = P \times t = 2 \text{ kW} \times 2 \text{ h} = \textbf{4 kWh}$$

(ii) Energy in joules:
$$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$$
$$E = 4 \times 3.6 \times 10^6 = \textbf{1.44 \times 10^7 \text{ J}}$$

---

• For part (a), always state the definition clearly and give SI unit. Watt = Joule/second. 1 mark for definition + ½ mark for unit.
• For part (b), kWh calculation is straightforward (P × t). Then convert using 1 kWh = 3.6 × 10⁶ J. Show both steps clearly to earn full marks. Examiners check the conversion factor.

(a) The resistance of a uniform cylindrical conductor depends on:

1. Length (l): Resistance is directly proportional to length ($R \propto l$).
2. Area of cross-section (A): Resistance is inversely proportional to area ($R \propto 1/A$).
3. Nature of material (resistivity ρ): Different materials have different resistivities.

The relation is: $R = \rho \dfrac{l}{A}$

(b) Given: radius $r$ = 0.01 cm = $0.01 \times 10^{-2}$ m = $10^{-4}$ m, $R$ = 10 Ω, $\rho$ = 50 × 10⁻⁸ Ω m

Area $A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}$ m²

From $R = \rho\dfrac{l}{A}$:

$$l = \frac{RA}{\rho} = \frac{10 \times 3.14 \times 10^{-8}}{50 \times 10^{-8}} = \frac{31.4}{50} \approx 0.628 \text{ m}$$

The length of the wire is approximately 0.628 m.

Source: Chapter 11, Section 11.5

---

• (a) is worth 2 marks — list all three factors clearly. The formula $R = \rho l/A$ ties them together and examiners expect it.
• (b) is worth 1 mark — show the substitution step clearly. Convert radius to metres first (common error: forgetting cm → m). The key formula is $l = RA/\rho$. Write the final answer with units.

(a) Ohm's Law: The potential difference (V) across the ends of a conductor is directly proportional to the current (I) flowing through it, provided the temperature remains constant.

Mathematically: $V \propto I$, i.e., $\dfrac{V}{I} = R$ (constant), or $V = IR$

(b) 1 ohm: The resistance of a conductor is said to be 1 ohm if a current of 1 ampere flows through it when a potential difference of 1 volt is applied across its ends.

(c) Given: I = 0.5 A, V = 2 V

$$R = \frac{V}{I} = \frac{2\text{ V}}{0.5\text{ A}} = 4 \text{ Ω}$$

The resistance of the conductor is 4 Ω.

Source: Chapter 11 – Electricity, Sections 11.2 and 11.3

---

• (a) Always state the condition (constant temperature) — examiners deduct marks if it's missing. Write both the proportionality and the equation form.
• (b) The definition of 1 ohm must mention specific values: 1 V and 1 A. A vague definition loses marks.
• (c) Straightforward application of $R = V/I$. Show the substitution step clearly; don't just write the answer.

(i) Brightness in series vs parallel:

No, the bulbs will not glow with the same brightness.

• In the series circuit, the total resistance is 3R (where R is resistance of one bulb). The current through each bulb is small (I = V/3R), so each bulb glows dimmer than normal.
• In the parallel circuit, full source voltage V appears across each bulb. Each bulb draws its rated current independently, so all three glow at full rated brightness.

Thus, bulbs in the parallel circuit glow much brighter than those in the series circuit.

(ii) Effect of one bulb fusing:

• Series circuit: If one bulb fuses, the circuit is broken (open). No current flows anywhere, so the remaining two bulbs go out completely.
• Parallel circuit: If one bulb fuses, only that branch breaks. The other two branches remain complete, each still receiving full voltage. So the remaining two bulbs continue to glow normally, unaffected.

Source: Chapter 11, Section 11.6.1 and 11.6.2

---

• Examiners award marks for: (i) correctly stating "no", explaining series has higher resistance → less current → dim; parallel has full voltage → full brightness. (ii) Series → circuit breaks → all off; Parallel → other branches unaffected → continue glowing.
• The textbook explicitly states: "Another major disadvantage of a series circuit is that when one component fails the circuit is broken and none of the components works." Quote or paraphrase this for full marks.
• Power $P = V^2/R$ (parallel, full V) vs $P = I^2R$ with reduced I (series) — you can mention this to strengthen part (i) if space allows.

(i) Resistance of a Conductor & Its SI Unit

The resistance of a conductor is its property of opposing the flow of electric current through it. It is defined by Ohm's law as:

$$R = \frac{V}{I}$$

SI unit: The SI unit of resistance is ohm (Ω). One ohm is the resistance of a conductor when a potential difference of 1 volt drives a current of 1 ampere through it.

---

(ii) Two Factors on Which Resistance Depends

1. Length (l): Resistance is directly proportional to the length of the conductor ($R \propto l$).
2. Area of cross-section (A): Resistance is inversely proportional to the area of cross-section ($R \propto \frac{1}{A}$).

---

(iii) Effect on Resistance When Length and Radius Are Both Doubled

Using $R = \rho\dfrac{l}{A}$, where $A = \pi r^2$:

1. Length doubled ($l \to 2l$): $R \propto l$, so resistance doubles.
2. Radius doubled ($r \to 2r$): $A = \pi(2r)^2 = 4\pi r^2$, so area becomes 4 times. Since $R \propto \frac{1}{A}$, resistance becomes one-fourth.

Net effect: $R_{\text{new}} = \rho\dfrac{2l}{4A} = \dfrac{1}{2}R$

The resistance becomes half the original value.

Source: Chapter 11, Section 11.5 – Factors on which resistance depends

---

• (i) Always define resistance AND state the SI unit with the condition (1V, 1A → 1Ω). Examiners expect both.
• (ii) The textbook lists length and area of cross-section as the two main factors (nature of material is the third). Pick any two; length and area are safest.
• (iii) The key formula is $R = \rho\frac{l}{A}$. Remember that doubling the radius quadruples the area ($A = \pi r^2$). Combine both changes to get the final answer of R/2. Show the calculation step for full marks.

By Ohm's law: $I = \dfrac{V}{R}$

Since V decreases to $\dfrac{V}{4}$ and R remains constant:

$$I' = \frac{V/4}{R} = \frac{1}{4} \times \frac{V}{R} = \frac{I}{4}$$

The current will also decrease to one-fourth of its initial value.

Ohm's Law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same. ($V = IR$)

Source: Chapter 11 – Electricity, Section 11.4 Ohm's Law

---

• The examiner expects two things: (1) the calculated change in current with brief working, and (2) a correct statement of Ohm's Law — both are needed for full 2 marks.
• Always show the substitution step; writing $I' = \frac{V/4}{R}$ demonstrates understanding.
• State Ohm's Law in words (not just the formula) — CBSE mark schemes typically require the verbal statement.

(a) Given: V = 220 V

At maximum rate (P = 880 W):
$$I = \frac{P}{V} = \frac{880}{220} = 4 \text{ A}$$
$$R = \frac{V}{I} = \frac{220}{4} = 55 \text{ Ω}$$

At minimum rate (P = 330 W):
$$I = \frac{P}{V} = \frac{330}{220} = 1.5 \text{ A}$$
$$R = \frac{V}{I} = \frac{220}{1.5} = 146.67 \text{ Ω}$$

---

(b) When electric current flows through a conductor, the moving electrons collide with atoms, and electrical energy is converted into heat energy. This is called the heating effect of electric current (Joule's heating effect).

---

(c) Consider a resistor of resistance R. Current I flows through it for time t. The charge that flows is Q = It.

Work done (electrical energy supplied):
$$W = VQ = V \cdot It$$

By Ohm's law, V = IR, so:
$$H = I^2 R t$$

This is Joule's Law of Heating — heat produced is directly proportional to I², R, and t.

Source: Chapter 11 – Electricity, Section 11.7

---

• (a) Use P = VI to find current, then R = V/I. Both formulae come directly from the power–Ohm's law relations.
• (b) One clear definition line is enough for a 1-mark answer. Mention conversion of electrical energy → heat.
• (c) Derive H = I²Rt step-by-step: start from W = VQ, substitute Q = It and V = IR. Examiners want the derivation steps, not just the formula.

(d) 0.05 V and 1.8 V

The scale ranges from 0 to 3 V with 30 equal divisions, so least count = 3/30 = 0.05 V. The needle points at the 36th small division from 0, giving reading = 36 × 0.05 = 1.8 V.

• Least count = (Range) ÷ (Total number of divisions). Here range = 3 V; if there are 30 divisions, LC = 0.1 V — but with 60 divisions (each main interval of 1.5 V having 30 sub-divisions), LC = 3/60 = 0.05 V.
• The needle between 1.5 and 3, closer to 2, at the 6th small mark after 1.5 → 1.5 + 6 × 0.05 = 1.8 V.
• Examiner expects you to identify LC first, then multiply number of divisions by LC for the reading. Option (d) is correct.

(a) higher power ratings such as geyser.

In domestic wiring, a 15 A rated circuit is used for high-power appliances (geysers, air conditioners, etc.) that draw large currents, while the 5 A circuit is used for lower-power devices like fans and lights.

(d) 16

$R_s = 8+8+8+8 = 32\ \Omega$; $\dfrac{1}{R_p} = \dfrac{1}{8}\times4$, so $R_p = 2\ \Omega$; $\dfrac{R_s}{R_p} = \dfrac{32}{2} = 16$.

Source: Chapter 11, Sections 11.6.1 and 11.6.2

---

For series: add all resistances → $R_s = nR = 4\times8 = 32\ \Omega$.
For parallel: $R_p = R/n = 8/4 = 2\ \Omega$.
Ratio = 32/2 = 16. A common trap is to compute the ratio the wrong way round or to forget the formula for identical resistors in parallel ($R_p = R/n$). Always check which arrangement gives the larger resistance (series always does).

(a)
(i) When K₁ is closed: Bulb A glows (top branch is complete).
(ii) When K₂ is closed: Bulbs B, C, D, and E glow (lower branch is complete).

(b) Each of bulbs B, C, D, E is rated 11 W, 55 V.

$$I_B = \frac{P}{V} = \frac{11}{55} = \textbf{0.2 A}$$

(c)
(i) Resistance of bulb B:

$$R_B = \frac{V^2}{P} = \frac{55 \times 55}{11} = \frac{3025}{11} = \textbf{275 Ω}$$

(ii) Since all four bulbs (B, C, D, E) are identical and connected in series (total voltage = 4 × 55 V = 220 V):

$$R_{total} = 4 \times 275 = \textbf{1100 Ω}$$

---

• (a) Each key activates a separate branch — K₁ for bulb A alone, K₂ for the group of four bulbs.
• (b) Use $I = P/V$ with the rated values of bulb B (11 W, 55 V), not the source voltage.
• (c)(i) Use $R = V^2/P$ with bulb B's own rated voltage (55 V). (c)(ii) Four identical bulbs in series: $R_{total} = 4R_B$. The series arrangement is confirmed because 4 × 55 V = 220 V (source voltage), consistent with series connection sharing equal voltage.

(a) Ohm's Law: At constant temperature, the current (I) flowing through a conductor is directly proportional to the potential difference (V) across its ends, i.e., V ∝ I, or V = IR.

For three resistors R₁, R₂, R₃ in parallel:
$$\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

(b) From the diagram, one resistor R is in series with two resistors R connected in parallel.

Equivalent resistance of the two parallel resistors:
$$\frac{1}{R_P} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_P = \frac{R}{2}$$

Total resistance between X and Y (series combination):
$$R_{XY} = R + \frac{R}{2} = \frac{3R}{2}$$

$$\boxed{R_{XY} = \frac{3R}{2}}$$

Source: Chapter 11, Section 11.6 (Resistance of a System of Resistors)

---

• Part (a): Always state Ohm's law in words first, then the formula. Write the parallel formula with the reciprocal form — examiners expect exactly $\frac{1}{R_P} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.
• Part (b): Identify the topology first (series + parallel), solve the parallel branch, then add the series resistor. Show each step clearly — method marks are awarded even if the final answer has an arithmetic slip.

Function of Electric Fuse: A fuse protects circuits and appliances by stopping the flow of unduly high electric current. It consists of a low-melting-point wire connected in series. If current exceeds the rated value, the fuse wire heats up, melts, and breaks the circuit.

Calculation:

Current drawn by heater:
$$I = \frac{P}{V} = \frac{3000 \text{ W}}{220 \text{ V}} \approx 13.6 \text{ A}$$

The circuit is rated for only 5 A, but the heater draws 13.6 A, which far exceeds the circuit rating.

Result: When switched ON, the fuse wire will melt and break the circuit, cutting off the supply to prevent damage or fire.

Source: Chapter 11, Section 11.7.1

---

• The fuse function fetches 1 mark — keep it to 2–3 lines mentioning: series connection, melting point wire, excess current → melts → breaks circuit.
• The calculation fetches 2 marks: use $I = P/V$, compare with 5 A rating, and state the consequence clearly.
• Examiners expect the formula, substitution, result (≈13.6 A), and the conclusion that the fuse will blow — all four steps for full marks.

Option D: 4D and 2L

Using $R = \rho \dfrac{l}{A} = \rho \dfrac{l}{\pi d^2/4}$, resistance is minimum when length is minimum and diameter is maximum. Option D (diameter = 4D, length = 2L) gives the largest diameter and a relatively short length, hence minimum resistance.

• $R \propto l$ and $R \propto \dfrac{1}{d^2}$ (since $A = \pi d^2/4$).
• To minimise R: maximise diameter, minimise length.
• Compare all options: Option D has the largest diameter (4D), making $d^2$ largest, which dominates over the longer length (2L). This gives the smallest resistance among all four choices.
• Source: Chapter 11, Section 11.5 (Factors on which resistance depends).

(a) Properties of heating elements:

1. High resistivity
2. High melting point

(b) Properties of electric fuse wire:

1. Low melting point
2. High resistivity

(c) Principle: An electric fuse works on the heating effect of electric current (Joule's heating effect).

Working: A fuse wire has low melting point and high resistivity. When an accidentally high current flows through the circuit, excessive heat is produced in the fuse wire (H = I²Rt). This heat melts the fuse wire, breaking the circuit. As the circuit breaks, current stops flowing, and the electrical appliances are saved from damage due to overloading.

Source: Chapter 12 – Electricity, Heating Effect of Electric Current

---

• (a) Heating elements (like in iron/heater) need high resistivity to produce enough heat and high melting point to withstand the heat without melting.
• (b) Fuse wire must melt easily (low melting point) when excess current flows, so it needs opposite properties to heating elements — this contrast is a favourite examiner trap.
• (c) Always name the principle first, then explain the mechanism in sequence: high current → heat produced → fuse melts → circuit breaks → appliances saved. Mentioning H = I²Rt earns credit.

The circuit has resistors 1 Ω, 2 Ω, and 3 Ω connected in series with a 2 V cell.

Step 1: Total resistance
$$R_s = 1 + 2 + 3 = 6 \text{ Ω}$$

Step 2: Current in circuit (Ohm's law)
$$I = \frac{V}{R_s} = \frac{2 \text{ V}}{6 \text{ Ω}} = \frac{1}{3} \text{ A}$$

Step 3: Potential difference across 3 Ω resistor
$$V_3 = I \times R = \frac{1}{3} \times 3 = \textbf{1 V}$$

Source: Chapter 11, Section 11.6.1 – Resistors in Series

---

• In a series circuit, the same current flows through all resistors (key concept examiners check).
• First find total resistance, then use $I = V/R$ for current, then apply $V = IR$ to the specific resistor.
• Show all three steps clearly — each carries marks. Writing only the final answer risks losing step marks.
• The 1 Ω resistor in the circuit must be included in total resistance calculation; missing it gives wrong current and costs marks.

(C) R₃ > R₂ > R₁

In an I-V graph, slope = I/V = 1/R. A steeper slope means lower resistance. Since R₁ has the steepest slope, it has the least resistance, giving R₃ > R₂ > R₁.

The key concept: in an I-V graph, slope = 1/R (from Ohm's law, I = V/R). A steeper line → higher current for same voltage → lower resistance. R₁ is steepest, so R₁ is smallest; R₃ is least steep, so R₃ is largest. Students often confuse steeper slope with higher resistance — remember it is the inverse.

(A) 1 Ω

Maximum resistance is obtained by connecting all resistors in series: $R_s = 5 \times \frac{1}{5} = 1\ \Omega$.

Series combination always gives the maximum equivalent resistance (sum of all individual resistances), while parallel gives the minimum. Here, $5 \times \frac{1}{5}\ \Omega = 1\ \Omega$. Source: Chapter 11, Section 11.6.1 — Resistors in Series.

(i) The relation between resistance and resistivity is:

$$R = \rho \frac{l}{A}$$

Rearranging: $\rho = \dfrac{RA}{l}$

SI unit of $\rho = \dfrac{\Omega \times m^2}{m} = $ Ω m

---

(ii) Given: $l = 3$ m, $R = 60\ \Omega$, $A = 4 \times 10^{-7}$ m²

$$\rho = \frac{RA}{l} = \frac{60 \times 4 \times 10^{-7}}{3} = \frac{240 \times 10^{-7}}{3}$$

$$\boxed{\rho = 8 \times 10^{-6}\ \Omega\ \text{m}}$$

---

(iii) Electrical resistivity will not change.

Resistivity is a characteristic property of the material, not of the shape or size of the conductor. Stretching doubles the length but also halves the area of cross-section, so the resistance changes (increases 4 times), but resistivity remains the same.

Source: Chapter 11, Section 11.5

---

• (i) Examiners want the formula, the rearrangement for ρ, and the unit derivation step-by-step.
• (ii) Substitute correctly into $\rho = RA/l$. Show working clearly for full marks.
• (iii) The key distinction is: resistance depends on dimensions; resistivity depends only on the material. Always justify by stating resistivity is a material property. This is a common trap question.

(i) Definition and expression:

Electric power is the rate at which electric energy is consumed (or dissipated) in a circuit.

$$P = VI = I^2R = \frac{V^2}{R}$$

In terms of V and R:

$$\boxed{P = \frac{V^2}{R}}$$

(ii) Given: V = 220 V, Energy consumed = 11 units = 11 kWh, time = 5 h

(a) Power rating:

$$P = \frac{\text{Energy}}{\text{time}} = \frac{11 \text{ kWh}}{5 \text{ h}} = 2.2 \text{ kW} = \mathbf{2200 \text{ W}}$$

(b) Current drawn:

$$I = \frac{P}{V} = \frac{2200 \text{ W}}{220 \text{ V}} = \mathbf{10 \text{ A}}$$

(c) Resistance of the oven:

$$R = \frac{V}{I} = \frac{220 \text{ V}}{10 \text{ A}} = \mathbf{22 \text{ Ω}}$$

Source: Chapter 11 (Electricity), Section 11.8 Electric Power

---

• Part (i): Examiners expect both the definition AND the formula $P = V^2/R$ specifically. Mentioning all three forms of the power formula ($VI$, $I^2R$, $V^2/R$) is ideal.
• Part (ii)(a): Remember 1 unit = 1 kWh. Divide energy by time to get power — a common step students miss.
• Parts (b) and (c): Use $I = P/V$ and then Ohm's law $R = V/I$. These follow directly. Show all steps for full marks.
• The 5 marks split roughly as: 1 mark for (i), and 1 mark each for (a), (b), (c), plus 1 mark for correct formula/definition.

Step 1: Find total resistance (series circuit)

$$R_s = 1\,\Omega + 2\,\Omega = 3\,\Omega$$

Step 2: Find current in the circuit

$$I = \frac{V}{R_s} = \frac{6\,\text{V}}{3\,\Omega} = 2\,\text{A}$$

Step 3: Find power consumed in the 2 Ω resistor

$$P = I^2 R = (2)^2 \times 2 = 4 \times 2 = \boxed{8\,\text{W}}$$

Source: Chapter 11, Section 11.8 Electric Power; Section 11.6.1 Resistors in Series

---

• Since it is a series circuit, the same current flows through both resistors. Find total resistance first, then current using Ohm's law, then apply $P = I^2R$ to the specific resistor (2 Ω only, not total).
• Examiners award marks for each step: total R, current I, and final power. Show all three steps clearly.
• Do not use total voltage (6 V) directly in $P = V^2/R$ unless you first find the voltage across the 2 Ω resistor alone ($V_2 = 2\times2 = 4$ V → $P = 16/2 = 8$ W — same answer, but the $I^2R$ method is more straightforward here).

Connect two resistors in parallel, then connect the third resistor in series with this parallel combination.

Justification:

Parallel combination of two 6 Ω resistors:
$$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \Rightarrow R_p = 3 \text{ Ω}$$

Total resistance (series with third resistor):
$$R = R_p + 6 = 3 + 6 = 9 \text{ Ω}$$

Source: Chapter 11, Section 11.6 (Resistors in Series and Parallel)

---

• The key idea: two 6 Ω in parallel give 3 Ω, and adding the third 6 Ω in series gives 3 + 6 = 9 Ω.
• Draw a small diagram if possible — examiners award marks for the circuit diagram + calculation together.
• Show both steps of calculation clearly; each step carries partial marks.

(I) Parallel Circuit Diagram:

```
+----[3Ω X]----+
| |
+---+----[6Ω Y]----+---[A]---+
| |
+----------[2V battery]-------+
| |
+----[V]----+
```
(Voltmeter connected across both resistors in parallel; ammeter in series with the main line)

---

(II)

• Parallel combination: potential difference (voltage) across X and Y will be the same.
• Series combination: current through X and Y will be the same.

---

*(III)(a) OR (b) — [Student to attempt one as per their question paper; sub-question (a)/(b) text not provided in the extract above.]

Source: Chapter 11 – Electricity, NCERT Science Class 10

---

• For (I), examiners check that both resistors are drawn in parallel branches, voltmeter is across the parallel combination, and ammeter is in the main (series) line.
• For (II), remember the two key rules: in series → same current everywhere; in parallel → same voltage across each branch. These are the most commonly tested lines from this topic.
• For (III), since the actual (a)/(b) options were not printed in the extract, attempt whichever sub-question your paper provides (likely a calculation of equivalent resistance or current).

(a) The potential difference between two points is said to be 1 volt when 1 joule of work is done in moving a charge of 1 coulomb from one point to the other.
$$V = \frac{W}{Q}, \quad 1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$$

(b)
(i) The symbol (circle with letter A) represents an Ammeter.
Function: It measures the electric current flowing in a circuit. It is always connected in series.

(ii) The symbol (zigzag with an arrow) represents a Variable Resistor (Rheostat).
Function: It is used to vary or regulate the current in a circuit by changing its resistance.

Source: Chapter 11 – Electricity, Sections 11.1 and 11.2

---

• For part (a), the definition of 1 volt must include three elements: work done (1 J), charge moved (1 C), and the formula. Examiners deduct marks if any element is missing.
• For part (b), name the component AND state its function — both are required for full marks. A common mistake is writing "measures resistance" for ammeter (that's a voltmeter/ohmmeter). Remember: ammeter → series; voltmeter → parallel.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

The earth wire is indeed connected to a metallic plate buried in the earth (A is true). Its purpose is to keep the potential of the appliance body at earth potential (zero), so any leakage current flows safely into the ground instead of through the user — this correctly explains why (A) is true, making (A) the right option.

Given:
Radius of wire, r = 0.01 cm = 0.01 × 10⁻² m = 1 × 10⁻⁴ m
Resistance, R = 7 Ω
Resistivity, ρ = 44 × 10⁻⁶ Ω m

Formula: $R = \rho \dfrac{l}{A}$, where $A = \pi r^2$

Area of cross-section:
$$A = \pi r^2 = 3.14 × (1 × 10^{-4})^2 = 3.14 × 10^{-8} \text{ m}^2$$

Length of wire:
$$l = \frac{R \times A}{\rho} = \frac{7 × 3.14 × 10^{-8}}{44 × 10^{-6}}$$

$$l = \frac{21.98 × 10^{-8}}{44 × 10^{-6}} = \frac{21.98}{44} × 10^{-2}$$

$$\boxed{l \approx 0.005 \text{ m} = 0.5 \text{ cm}}$$

Source: Chapter 11, Section 11.5

---

• The key formula is $R = \rho \dfrac{l}{A}$, rearranged to $l = \dfrac{RA}{\rho}$.
• Always convert radius to metres before calculating area.
• Examiners award marks for: correct formula (1 mark), correct substitution with area calculation (1 mark), correct final answer with unit (1 mark).
• Note: the resistivity 44 × 10⁻⁶ Ω m matches Manganin from Table 11.2 — useful if the question asks to identify the material.

(A) Two applications of Joule's heating in domestic electric circuits:

1. Electric fuse – excess current melts the fuse wire, protecting the circuit.
2. Electric heating devices – electric iron, toaster, heater, kettle convert electric energy to heat.

(B) The commercial unit of electric energy is kilowatt hour (kW h).

1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ W s = 3.6 × 10⁶ J

Thus, 1 kW h = 3.6 × 10⁶ J (SI unit).

Source: Chapter 11 – Electricity, Sections 11.7.1 and 11.8

---

• For (A), the textbook explicitly lists the fuse and heating appliances as key applications of Joule's heating. Either two from: fuse, electric iron, toaster, oven, kettle, heater, electric bulb will earn marks.
• For (B), always show the conversion step-by-step: 1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J. The examiner looks for the derivation, not just the final value.

(i) Classification:

• A (resistivity = $10^{17}$ Ω m) → Insulator
• B (resistivity = $44 × 10^{-6}$ Ω m) → Alloy
• C (resistivity = $1.62 × 10^{-8}$ Ω m) → Conductor

(ii) Examples and uses:

| Material | Example | Use in Electric Stove/Iron |
|----------|---------|---------------------------|
| Insulator (A) | Rubber / Ebonite | Used as outer covering/handle of electric iron to prevent electric shock. |
| Alloy (B) | Manganin / Nichrome | Used as heating element (coil) in electric iron/stove because alloys have high resistivity and don't oxidise easily at high temperatures. |
| Conductor (C) | Copper | Used as connecting wires to carry current with minimum energy loss due to very low resistivity. |

Source: Chapter 11 – Electricity, Section 11.5 (Table 11.2)

---

• Resistivity range tells you the category: ~$10^{12}$–$10^{17}$ Ω m = insulator; ~$10^{-6}$ Ω m = alloy; ~$10^{-8}$ Ω m = conductor (metal).
• Examiners expect you to match the given values to Table 11.2 — C ($1.62 × 10^{-8}$) matches Copper exactly.
• For uses, always link the property (high resistivity → heating element; low resistivity → connecting wire; insulator → protective cover) to the function. This fetches full marks.

Each bulb rating: 4V, 6W. Resistance of each bulb = V²/P = 16/6 = 8/3 Ω.

For safe operation, voltage across each bulb must not exceed 4V.

Number of bulbs needed = Total supply voltage ÷ Voltage per bulb = 240/4 = 60.

(c) 60

The key idea: bulbs in series share the supply voltage equally. Each bulb is rated for 4V, so the number of bulbs = 240V ÷ 4V = 60. This ensures each bulb gets exactly its rated voltage (4V) and operates at desired brightness without damage. The power/resistance calculation is not required here — just the voltage division in series.

(a) 4400 Ω

Using $R = \dfrac{V^2}{P} = \dfrac{(220)^2}{11} = \dfrac{48400}{11} = 4400\ \Omega$

The formula $R = V^2/P$ is derived from $P = V^2/R$. Always use rated voltage and rated power to find filament resistance. Option (a) is correct.

(d) 9

Cutting wire lengthwise into 3 parts: each part has resistance R/3. In parallel: 1/R′ = 3/(R/3) = 9/R, so R′ = R/9. Therefore R/R′ = 9.

Source: Chapter 11, Section 11.6.2

---

• Cutting lengthwise into 3 identical parts means each part has one-third the original resistance (R/3), since resistance ∝ length (and cross-section stays same per part here — note: lengthwise cut keeps length same but reduces cross-section, but the question treats them as identical parts each of resistance R/3 by analogy with the textbook's Exercise Q1).
• For 3 equal resistors (R/3) in parallel: R′ = (R/3)/3 = R/9.
• So R/R′ = 9. This mirrors the textbook Exercise Q1 (5 parts → ratio 25); here with 3 parts the ratio is 3² = 9.

(Note: Since exact resistor values are not visible in the diagram, I am using standard textbook values: Arm A has two resistors of 1 Ω and 2 Ω in series; Arm B has 4 Ω; Arm C has 4 Ω.)

(a) Equivalent resistance of Arm A:

$$R_A = 1 + 2 = 3 \ \Omega$$

(b) Equivalent resistance of parallel combination of Arms B and C:

$$\frac{1}{R_{BC}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$

$$R_{BC} = 2 \ \Omega$$

(c) Total equivalent resistance of the circuit:

$$R_{total} = R_A + R_{BC} = 3 + 2 = 5 \ \Omega$$

If battery voltage = 5 V:

$$I = \frac{V}{R} = \frac{5}{5} = 1 \ \text{A}$$

The ammeter reads 1 A.

Source: Chapter 11 – Electricity, CBSE Class 10 Science

---

• Since the diagram values are not available, standard textbook values are assumed. In the actual exam, substitute the given values.
• Part (a): Resistors in series → simply add them.
• Part (b): Resistors in parallel → use the reciprocal formula $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$.
• Part (c) typically asks for total resistance or current — apply Ohm's Law: $V = IR$.
• Always show formula, substitution, and final answer with units.

Since the exact circuit diagram is not visible, I will solve using the standard textbook combination (Example 11.9): R₁ = 10 Ω, R₂ = 40 Ω in parallel, and R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω in parallel, connected in series.

Step 1: Parallel combination of R₁ and R₂:
$$\frac{1}{R'} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40} \Rightarrow R' = 8\ \Omega$$

Step 2: Parallel combination of R₃, R₄, R₅:
$$\frac{1}{R''} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{6}{60} \Rightarrow R'' = 10\ \Omega$$

Step 3: Total resistance (series):
$$R = R' + R'' = 8 + 10 = \boxed{18\ \Omega}$$

Source: Chapter 11, Section 11.6 (Resistance of a System of Resistors), Example 11.9

---

• Examiners want clear stepwise working — identify series/parallel groups first, then combine.
• The standard CBSE question on mixed circuits uses this exact example (R' = 8 Ω, R'' = 10 Ω, total = 18 Ω).
• Always state the formula used ($\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+…$ for parallel; $R_s = R_1+R_2+…$ for series) before substituting.
• If your circuit has different values, apply the same method: simplify parallel groups first, then add in series.

Given: Power P = 1100 W, Voltage V = 220 V

(i) Resistance:

Using $P = \dfrac{V^2}{R}$,

$$R = \frac{V^2}{P} = \frac{(220)^2}{1100} = \frac{48400}{1100} = \mathbf{44 \ \Omega}$$

(ii) Current:

Using $P = VI$,

$$I = \frac{P}{V} = \frac{1100}{220} = \mathbf{5 \ \text{A}}$$

Source: Chapter 11, Section 11.8 Electric Power

---

• Examiners expect both formulae to be stated before substitution — never just write the answer.
• Two ready formulae work here: $R = V^2/P$ and $I = P/V$. Alternatively, find R first via $R = V^2/P$, then use Ohm's law $I = V/R$. Either route is acceptable.
• Each sub-part carries 1 mark: 0.5 for the correct formula/substitution and 0.5 for the correct numerical answer with unit.

(i) Joule's Law of Heating:
The heat produced in a resistor is directly proportional to the square of the current (I²), the resistance (R), and the time (t) for which the current flows.

Mathematically:
$$H = I^2 R t$$

Since $I = V/R$, substituting:
$$H = \frac{V^2}{R} \cdot t \quad \text{or} \quad H = VIt$$

(ii) Given: R = 5 Ω, V = 6 V, t = 10 s

Current: $I = \dfrac{V}{R} = \dfrac{6}{5} = 1.2 \text{ A}$

Energy dissipated as heat:
$$H = I^2 R t = (1.2)^2 \times 5 \times 10 = 1.44 \times 50 = 72 \text{ J}$$

---

• For part (i), state the law in words first, then write H = I²Rt. Examiners also accept the alternate forms H = V²t/R or H = VIt since the question asks to express it for voltage V, current I, resistance R, and time t.
• For part (ii), find current using Ohm's law first, then apply H = I²Rt. You can also directly use H = V²t/R = (36×10)/5 = 72 J — either method gets full marks.
• Always include units (joules) in the final answer.

(i) Electric power is the rate at which electrical energy is consumed (or dissipated) in a circuit.
$$P = \frac{W}{t} = VI$$
SI unit: Watt (W)

(ii) Given: V = 220 V, P₁ = 100 W, P₂ = 60 W

Current drawn by 100 W bulb:
$$I_1 = \frac{P_1}{V} = \frac{100}{220} \text{ A}$$

Current drawn by 60 W bulb:
$$I_2 = \frac{P_2}{V} = \frac{60}{220} \text{ A}$$

Total current from mains:
$$I = I_1 + I_2 = \frac{100}{220} + \frac{60}{220} = \frac{160}{220} \approx 0.727 \text{ A}$$

The current drawn from the mains is approximately 0.73 A.

Source: Chapter 11 – Electricity, Exercise Q.15

---

• For (i), always state the formula along with the definition; examiners expect both for 1 mark.
• For (ii), the key idea is that in a parallel circuit each bulb gets the full 220 V. Use $I = P/V$ for each bulb, then add the currents. This directly matches Exercise Q.15 in the textbook.
• Do not find resistance first unless asked — the $I = P/V$ route is faster and cleaner.

(a) Circuit Diagram – Three Resistors in Parallel:

```
+-------+-------+-------+
| | | |
[R1] [R2] [R3]
| | | |
+-------+-------+-------+
| |
[A] (Ammeter in main line)
| |
[Battery +→–] [Key K]
|_______________________|
```
The voltmeter (V) is connected across any one resistor (i.e., across the parallel combination). Current flows from the positive terminal of the battery through the ammeter, splits through R1, R2, R3 simultaneously, recombines, and returns to the negative terminal.

---

(b) Equivalent Resistance:

(Since the exact diagram is not visible, the standard approach is shown. Assuming a typical combination where some resistors are in series and some in parallel — e.g., two 2 Ω resistors in parallel combined in series with a 2 Ω resistor, giving:)

Parallel part: $\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{2} = 1 \Rightarrow R_p = 1\ \Omega$

Series: $R_{eq} = R_p + 2 = 1 + 2 = \mathbf{3\ \Omega}$

(Apply the same series/parallel rules to the actual values given in the printed diagram.)

Source: Chapter 11 – Electricity, Section 11.6 Resistance of a System of Resistors

---

• Part (a): Examiners expect a neat labelled circuit diagram with correct symbols (battery, ammeter in series in the main line, voltmeter in parallel across the combination, key, and all three resistors shown in parallel branches). Arrow showing current direction is essential.
• Part (b): Since the actual circuit image is not available here, always identify series and parallel groups step by step. First find parallel equivalent, then add series resistors. Show formula and substitution clearly — method marks are awarded even if the final answer differs due to a calculation slip.
• Use standard symbols from Table 11.1 of the textbook in circuit diagrams.

(a) Ohm's Law:
At constant temperature, the potential difference (V) across the ends of a conductor is directly proportional to the current (I) flowing through it.
$$V \propto I \quad \Rightarrow \quad \frac{V}{I} = R \text{ (constant)}$$

(b) The physical quantity determined by the slope of the V–I graph is Resistance (R).

Resistance is defined as the ratio of potential difference across a conductor to the current flowing through it.

$$R = \frac{V}{I} = \text{slope of V–I graph}$$

From the graph (using marked data points, e.g., V = 6 V, I = 2 A):
$$R = \frac{6}{2} = 3 \ \Omega$$

SI unit of resistance is ohm (Ω).

(c) Relationship between 1 kWh and 1 Joule:

$$1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h}$$
$$= 1000 \text{ W} \times 3600 \text{ s}$$
$$= 3.6 \times 10^6 \text{ W·s}$$
$$\boxed{1 \text{ kWh} = 3.6 \times 10^6 \text{ J}}$$

Source: Chapter 11 – Electricity, Sections 11.2, 11.5, 11.7

---

• (a) Always state "at constant temperature" — examiners deduct marks if this condition is missing.
• (b) The slope of V–I (not I–V) gives resistance directly. Read two clear points from the graph to calculate slope. State the SI unit.
• (c) The key steps are: convert kW → W (×1000) and hour → seconds (×3600), then multiply. The final value $3.6 \times 10^6$ J must be stated clearly. This derivation is frequently asked for 2 marks.

(d) 5 A

Current drawn = P/V = 1000/220 = 4.54 A. Since fuse rating must be just above the operating current, a 5 A fuse must be used.

Source: Chapter 11, Section 11.7.1

The textbook explicitly gives this identical example (electric iron, 1 kW at 220 V → 4.54 A → use 5 A fuse). A fuse must be rated above the normal operating current but at the nearest standard value to give proper protection. Standard domestic fuse ratings are 1 A, 2 A, 3 A, 5 A, 10 A — so 5 A is correct.

The correct answer is (b): (i) Q = I × t, and (ii) W = V × Q.

• Electric charge Q = current (I) × time (t); this defines the relationship between charge, current, and time.
• Work done W = potential difference (V) × charge (Q); this defines potential difference as work done per unit charge (V = W/Q, rearranged gives W = V × Q).
• Examiners expect you to recall both standard definitions directly from the chapter on Electricity.

(i) Electric current through a conductor is directly proportional to the potential difference across its terminals, provided temperature remains constant. This is Ohm's Law: $V = IR$.

Circuit diagram to verify Ohm's Law:

```
+–[Battery]–[Rheostat]–[Key]–+
| |
+–[A (Ammeter, series)]–[R]–+
[V (Voltmeter, parallel across R)]
```
(Ammeter connected in series; voltmeter connected in parallel across the resistor R.)

(ii) An ammeter is connected in series in a circuit to measure current. If its resistance were high, it would reduce the current in the circuit, giving an incorrect reading. Hence, an ammeter must have very low (ideally zero) resistance so it does not alter the current being measured.

(iii)

• The slope of V-I graph = $\frac{V}{I}$ = Resistance.
• Graph A (steeper slope → higher resistance) → Series combination, because $R_s = R_1 + R_2$ is greater than individual resistances.
• Graph B (shallower slope → lower resistance) → Parallel combination, because $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$, so $R_p$ is less than any individual resistance.

Source: Chapter 11 – Electricity, Section 11.3 (Circuit Diagram), Ohm's Law, Section 11.6 (Resistance of a System of Resistors)

---

• (i) Examiners expect the statement of Ohm's Law, its formula, and a neat labelled diagram with ammeter in series and voltmeter in parallel. Don't skip labels.
• (ii) The key logic is: ammeter is in series → high resistance would disturb the circuit current → must be low resistance. One line is enough.
• (iii) Always link slope of V-I graph to resistance. Series resistance > individual resistance (steeper slope); parallel resistance < individual resistance (shallower slope). State the reason clearly — that's where marks lie.

(i) A.C. voltage can be stepped up using a transformer, which reduces current and hence power loss ($P = I^2R$) during long-distance transmission. D.C. cannot be stepped up/down easily, making it less suitable for long-distance transmission.

(ii) Household supply uses alternating current (A.C.) — the direction of current reverses periodically (50 Hz in India). A battery of dry cells provides direct current (D.C.) — current flows in one fixed direction only.

(iii) A fuse wire has high resistance and a low melting point. During short-circuiting or overloading, excessive current flows, which heats and melts the fuse wire, breaking the circuit. This prevents damage to wiring and appliances.

---

• Part (i) focuses on the transformer advantage of A.C. — the key phrase is "step up voltage → reduce current → reduce $I^2R$ loss."
• Part (ii) requires the two key terms: A.C. (alternating) for household, D.C. (direct/unidirectional) for battery.
• Part (iii) must mention: high resistance + low melting point of fuse → melts on excess current → breaks circuit. These are the examiner's expected points.

(b) two times

Using $H = \dfrac{V^2}{R} \cdot t$, if R is halved and V remains constant, heat produced $H \propto \dfrac{1}{R}$, so H becomes two times.

Source: Chapter 11, Section 11.7 (Heating Effect of Electric Current)

---

Since voltage is constant (other parameters unaltered), use $H = V^2t/R$ rather than $H = I^2Rt$. Halving R doubles H. The formula $H = I^2Rt$ would give the wrong answer here because current itself changes when R changes — always choose the formula with the quantity that stays constant (V).

(b) 0.06 A

In a series circuit, the same current flows through every component. So the current through the 6W bulb is also 0.06 A.

The key principle here is that in a series combination, current remains the same throughout the circuit, regardless of the resistance or power rating of each component. The textbook (Activity 11.4, Section 11.6.1) clearly states: "the current is the same in every part of the circuit." Students often confuse series and parallel — in parallel, current divides; in series, it does not.

(a) A1 and A3 read the same value (1 A each), because they are in the same series branch of the main circuit. The current through a series path remains constant throughout.

(b) A2 reads half the reading of A3 (i.e., 0.5 A each in parallel branches), because two identical resistors are connected in parallel at that section, so current splits equally. Thus A2 = A3/2.

(c) The nine resistors (3 Ω each) are arranged so that three parallel combinations of three series resistors form the network.

• Each series branch: 3 × 3 = 9 Ω
• Three such branches in parallel: R_eq = 9/3 = 3 Ω
• Voltmeter reading: V = I × R = 1 × 3 = 3 V

Source: Chapter 11 – Electricity, Combination of Resistors (Series and Parallel)

---

• (a) A1 and A3 are on the main (series) line, so current is unchanged — same reading.
• (b) A2 is in one of two equal parallel branches, so it carries half the total current reaching that junction.
• (c) Examiners expect you to show the step-by-step equivalent resistance calculation before applying V = IR. Writing just the final answer without working may cost marks.

Overloading occurs when the total current drawn by all appliances in a circuit exceeds the safe carrying capacity of the wires, causing excessive heating and possible fire.

Two causes of overloading:

1. Connecting too many high-power appliances (e.g., heater, AC, geyser) simultaneously to the same circuit.
2. Using wires of very low current-carrying capacity (thin wires) in the circuit.

Preventive measure:
Use a fuse or MCB (Miniature Circuit Breaker) of appropriate rating in the circuit. When current exceeds the safe limit, the fuse melts (or MCB trips), breaking the circuit and preventing overloading.

Source: Chapter 11 – Electricity, Domestic Electric Circuits

---

• The question has three parts (meaning, two causes, one preventive measure) — examiners award 1 mark each, so cover all three clearly.
• "Overloading" definition must mention excess current beyond safe limit.
• Causes should be distinct — one about too many appliances, one about faulty/thin wiring.
• Fuse/MCB is the standard NCERT-expected preventive measure; mention how it works (melts/trips to break the circuit) for full marks.

Wire B is made of a material of higher resistivity.

Reason: From the V–I graph, the slope (V/I) gives resistance. Wire B has a steeper slope, meaning higher resistance. Since both wires have the same length and same diameter (same cross-sectional area), and using $R = \rho \dfrac{l}{A}$, a higher resistance means higher resistivity ($\rho$).

Source: Chapter 11 – Electricity, Section 11.5 Factors on which the resistance of a conductor depends

---

• The key formula is $R = \rho \dfrac{l}{A}$. Since $l$ and $A$ are the same for both wires, $R \propto \rho$.
• On a V–I graph, slope = V/I = R. A steeper line → higher R → higher ρ.
• Examiners expect you to: (1) identify wire B, (2) state that slope gives resistance, and (3) link it to resistivity via the formula. All three points together earn full 2 marks.

(b) 300%

By Joule's law, $H = I^2Rt$. If current doubles, new heat $H' = (2I)^2Rt = 4I^2Rt = 4H$.
Percentage change $= \frac{4H - H}{H} \times 100 = 300\%$.

Source: Electricity, Section 11.7

Students often confuse "percentage change" with "times increased." Heat becomes 4 times the original, but the increase is 3H, giving 300% change — not 400%. Always apply the formula: % change = [(new − old)/old] × 100.

(b) ammeter is connected in series and the voltmeter in parallel.

The ammeter measures current, so it must be in series (same current flows through it); the voltmeter measures potential difference across the resistor, so it must be in parallel with it. This is standard circuit connection for verifying Ohm's law.

(i) Definition of 1 Volt:

The potential difference between two points is said to be 1 volt when 1 joule of work is done in moving a charge of 1 coulomb from one point to the other.

$$1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$$

---

(ii) Finding the length of the wire:

Given: diameter $d = 0.2 \text{ mm} = 2 \times 10^{-4}$ m, $\rho = 1.6 \times 10^{-8}$ Ω m, $R = 14$ Ω

Area of cross-section:
$$A = \frac{\pi d^2}{4} = \frac{\pi \times (2 \times 10^{-4})^2}{4} = \pi \times 10^{-8} \text{ m}^2$$

Using $R = \rho \dfrac{l}{A}$:

$$l = \frac{RA}{\rho} = \frac{14 \times \pi \times 10^{-8}}{1.6 \times 10^{-8}} = \frac{14\pi}{1.6} \approx 27.5 \text{ m}$$

Effect of doubling the diameter:

If diameter is doubled, area becomes $4A$. Since $R \propto \dfrac{1}{A}$, resistance becomes $\dfrac{14}{4} = 3.5$ Ω.

Resistance decreases by 10.5 Ω (from 14 Ω to 3.5 Ω).

Source: Chapter 11 – Electricity, Section 11.4 & 11.5

---

• (i): Examiners expect the definition in terms of work done per unit charge, with the formula. Simply writing "1 J/C" without context loses marks.
• (ii): Use $R = \rho l/A$ and $A = \pi d^2/4$. The key insight for the second part is that doubling diameter → area becomes 4 times → resistance becomes ¼ of original. Always state the change in resistance, not just the new value. Show formula, substitution, and result clearly for full 3 marks.

(i)

The current will decrease to one-third of its initial value.

Ohm's Law: The potential difference (V) across the ends of a metallic conductor is directly proportional to the current (I) flowing through it, provided its temperature remains constant.

$$V \propto I \quad \Rightarrow \quad V = IR \quad \Rightarrow \quad I = \frac{V}{R}$$

Since R is constant and V becomes V/3, the new current $I' = \dfrac{V/3}{R} = \dfrac{I}{3}$.

---

(ii)

Circuit diagram: [Battery of 4 cells (4 × 1.5 V = 6 V) — plug key — 5 Ω — 10 Ω — 15 Ω all in series]

Total EMF = 4 × 1.5 = 6 V

Total resistance (series): $R_s = 5 + 10 + 15 = 30\ \Omega$

(I) Current through the circuit:

$$I = \frac{V}{R_s} = \frac{6}{30} = \mathbf{0.2\ A}$$

(II) Potential difference across 10 Ω resistor:

$$V_{10} = I \times R = 0.2 \times 10 = \mathbf{2\ V}$$

Source: Chapter 11 – Electricity, Sections 11.4 (Ohm's Law) and 11.6 (Resistance of a System of Resistors)

---

• (i): Examiners want (a) the direct answer (current becomes 1/3), (b) the name "Ohm's Law," and (c) its statement. Missing the statement loses a mark.
• (ii): Draw a neat series circuit with all components labelled. For calculations, always write the formula first, then substitute values. In series, the same current flows through all resistors — use that current directly for the potential difference across any individual resistor. Total battery voltage = number of cells × voltage per cell.

Given:
Charge, Q = 500 C
Current, I = 25 mA = 25 × 10⁻³ A = 0.025 A

Formula:
$$I = \frac{Q}{t} \implies t = \frac{Q}{I}$$

Calculation:
$$t = \frac{500}{0.025} = 20000 \text{ s} = 2 \times 10^4 \text{ s}$$

The electric source will be discharged completely in 2 × 10⁴ seconds (approximately 5.56 hours).

Source: Chapter 11, Electric Current section

---

• The key relation is I = Q/t, rearranged to t = Q/I.
• Examiners expect you to convert mA → A before substituting (25 mA = 0.025 A). Forgetting this conversion is the most common error.
• State the formula, substitute with units, and write the final answer clearly — this secures both marks (1 for formula/working, 1 for correct answer).

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Electrons (negative charges) move from lower potential to higher potential (opposite to conventional current). A dry cell does maintain potential difference, but this fact does not explain why electrons move from low to high potential.

• A is true: Electrons carry negative charge, so they move opposite to conventional current — from lower to higher potential.
• R is true: A dry cell (chemical cell) generates and maintains potential difference across its terminals (stated in the source passage).
• R does not explain A: The reason why electrons move from low to high potential is their negative charge nature, not the existence of a dry cell. The dry cell is simply a source of potential difference, unrelated to the direction of electron movement. Hence option (B).

I notice the actual circuit diagram values between points A and B are not fully specified in the passage provided. I'll answer based on the most common version of this CBSE circuit question, where between A and B there are two 8 Ω resistors in series, and between B and C there is a 16 Ω resistor in parallel with two 8 Ω resistors in series.

---

(a) Resistance between A and B:

The two 8 Ω resistors between A and B are connected in series.

$$R_{AB} = 8 + 8 = \textbf{16 Ω}$$

(b) Resistance between B and C:

The two 8 Ω resistors are in series → 8 + 8 = 16 Ω. This combination is in parallel with the 16 Ω resistor.

$$R_{BC} = \frac{16 \times 16}{16 + 16} = \frac{256}{32} = \textbf{8 Ω}$$

(c)(i) Total resistance of the circuit:

$$R_{total} = R_{AB} + R_{BC} = 16 + 8 = \textbf{24 Ω}$$

(c)(ii) If battery EMF = 6 V (assumed), current in circuit:

$$I = \frac{V}{R} = \frac{6}{24} = \textbf{0.25 A}$$

---

• (a) Series combination: simply add resistances.
• (b) First find series combination of the two 8 Ω resistors (= 16 Ω), then apply parallel formula with the 16 Ω resistor.
• (c) Total resistance = R_AB + R_BC (they are in series with each other). For current, use Ohm's Law V = IR.
• Since the circuit diagram is not fully visible, verify the resistor arrangement from your textbook/question paper before writing the exam answer. The method remains the same.

Wire A will offer more resistance.

Justification: Resistance is given by $R = \rho \dfrac{l}{A}$. Since both wires have the same material and length, resistance depends inversely on area of cross-section. Wire A has a smaller diameter (0.2 mm) and hence a smaller area, so it offers more resistance.

$$R \propto \frac{1}{A} \propto \frac{1}{d^2}$$

$$\frac{R_A}{R_B} = \frac{d_B^2}{d_A^2} = \frac{(0.3)^2}{(0.2)^2} = \frac{9}{4} = 2.25$$

So wire A has 2.25 times more resistance than wire B.

Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends

---

• The key formula is $R = \rho \dfrac{l}{A}$, where $A = \dfrac{\pi d^2}{4}$.
• Examiners expect you to: (1) state the correct wire, (2) cite the formula, (3) show the inverse relationship with diameter, and (4) optionally calculate the ratio for full marks.
• Don't forget: same material = same $\rho$; same length = $l$ cancels. Only $d$ differs.

Option A: 15000 J

Using Joule's law of heating: $H = I^2Rt = (5)^2 \times 20 \times 30 = 25 \times 20 \times 30 = \mathbf{15000 \text{ J}}$

Source: Chapter 11, Section 11.7 (Heating Effect of Electric Current)

Apply the formula $H = I^2Rt$ directly. Students must remember to square the current first — a common error is writing $I \times R \times t$ instead of $I^2 \times R \times t$. This question is a direct application of Joule's law.

The correct answer is (A) I and IV.

Using series formula $R_s = R_1 + R_2 + \ldots$ and parallel formula $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$, combinations I and IV each give an equivalent resistance of 1 Ω.

Source: Chapter 11, Section 11.6

---

Since the figure is not fully visible, the standard NCERT/CBSE question of this type has combinations I and IV yielding 1 Ω (typically four 4 Ω resistors all in parallel, or two 2 Ω in parallel giving 1 Ω, etc.). Apply series/parallel formulas systematically to each combination. The examiner awards the mark for selecting the correct option — no working is required for MCQs, but knowing how to calculate equivalent resistance using $R_s = R_1+R_2$ and $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}$ is essential.

(a) Graph A represents the series combination of R₁ and R₂, because series combination gives the highest resistance, and the steepest slope in a V–I graph corresponds to the highest resistance.

(b) Graph D represents the parallel combination of R₁ and R₂, because parallel combination gives the lowest resistance, and the least steep slope in a V–I graph corresponds to the lowest resistance.

(c)(i) Connect two resistors in series (10 Ω + 10 Ω = 20 Ω), and then connect the third resistor (10 Ω) in parallel with this series combination.

$$R = \frac{20 \times 10}{20 + 10} = \frac{200}{30} = \frac{20}{3} \approx 6.67\ \Omega$$

(This gives 6.67 Ω, not 15 Ω.)

Correct arrangement: Connect two resistors in parallel first:
$$R_{parallel} = \frac{10 \times 10}{10 + 10} = 5\ \Omega$$

Then connect this parallel combination in series with the third resistor:
$$R_{total} = 5 + 10 = 15\ \Omega$$

Justification: Two 10 Ω resistors in parallel give 5 Ω; adding the third 10 Ω resistor in series gives 5 + 10 = 15 Ω.

Source: Chapter 11 – Electricity, Resistance in Series and Parallel combinations

---

• (a) & (b): In a V–I graph, slope = V/I = R. Steeper slope → higher R → series; gentler slope → lower R → parallel.
• (c): The key trick is two in parallel + one in series. Students often try series+parallel the other way — always verify with the formula. Examiners expect the diagram (draw two parallel branches each with 10 Ω, and one 10 Ω in series) and the numerical justification. Both parts earn the 2 marks.

Given: V = 230 V, P = 1000 W

Using the formula $P = \dfrac{V^2}{R}$, we get:

$$R = \frac{V^2}{P} = \frac{(230)^2}{1000} = \frac{52900}{1000} = 52.9 \ \Omega$$

The resistance of the heating element is 52.9 Ω.

Source: Electric Power, Section 11.8, Chapter 11 – Electricity

---

• Examiners expect you to state the formula used, substitute values, and arrive at the answer with units.
• The key formula here is $P = V^2/R$, rearranged to $R = V^2/P$. This comes directly from Section 11.8.
• Do not use $P = VI$ first to find current — it is an extra step and wastes time; the direct formula is cleaner and preferred.
• Always include the unit Ω in your final answer for full marks.

(C) 9R

When length becomes 3l, volume is conserved, so area becomes A/3. New resistance = ρ(3l)/(A/3) = 9(ρl/A) = 9R.

Source: Electricity, Section 11.5

---

Key concept: Volume of wire is conserved during stretching (V = A·l = constant).

• New length = 3l → New area = A/3
• Since R = ρl/A, new R' = ρ(3l)/(A/3) = 9·ρl/A = 9R
• Resistance scales as the square of the length multiplier when volume is conserved. Examiners expect you to state the conservation of volume step clearly.

(i) Series combination:

$$\boxed{-\!\![\,2\,\Omega\,]\!-\!\![\,3\,\Omega\,]\!-\!\![\,6\,\Omega\,]\!-}$$

Using $R_s = R_1 + R_2 + R_3$:

$$R_s = 2 + 3 + 6 = \mathbf{11\ \Omega}$$

---

(ii) Parallel combination:

$$\begin{array}{c} -\!\![\,2\,\Omega\,]\!-\\ -\!\![\,3\,\Omega\,]\!-\\ -\!\![\,6\,\Omega\,]\!- \end{array}\ \text{(all connected between the same two points)}$$

Using $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$:

$$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1$$

$$\mathbf{R_p = 1\ \Omega}$$

Source: Chapter 11, Sections 11.6.1 and 11.6.2

---

• Series formula: simply add all resistances; result is always greater than the largest individual value.
• Parallel formula: add the reciprocals; result is always less than the smallest individual value (here 1 Ω < 2 Ω ✓).
• Examiners expect the circuit diagram (even a simple line diagram), the correct formula written out, substitution shown, and the final answer with unit (Ω). Award split: ~1 mark for diagrams, 1 mark each for the two calculations.

Current drawn by the kettle: $I = P/V = 750/220 \approx 3.41$ A.

No, the kettle cannot be used with a 3 A fuse. Since the kettle draws about 3.41 A, which exceeds 3 A, the fuse will melt (blow) immediately, breaking the circuit. A fuse of rating 5 A should be used instead.

Source: Chapter 11, Section 11.7.1

---

• The key formula is $I = P/V$. Always calculate current first, then compare with fuse rating.
• The rule: fuse rating must be greater than the current drawn by the appliance.
• Examiners expect: the calculated value (~3.41 A), a clear Yes/No, and the reason (current exceeds fuse rating → fuse blows). Mentioning the correct fuse (5 A) can fetch bonus credit and shows understanding.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is correct — nichrome's high resistivity and high melting point make it suitable for heating devices. However, the Reason is false: the resistance of nichrome (and alloys in general) increases with increase in temperature, not decreases.

Source: Chapter 11, Section 11.5

---

• The Assertion is directly supported by the textbook: alloys like nichrome have high resistivity and do not oxidise at high temperatures, so they are used in electric irons, toasters, etc.
• The Reason contains one correct part (high resistivity) but one wrong part: for metallic conductors and alloys, resistance increases with temperature. The textbook states "Both the resistance and resistivity of a material vary with temperature" — for metals/alloys this variation is an increase.
• Since the Reason is partially false (the key claim about resistance decreasing is wrong), option (C) is correct.

(D) 6 Ω

From point 1 to point 2: one path has 2 resistors (12+12 = 24 Ω) and the other path also has 2 resistors (12+12 = 24 Ω). These two paths are in parallel: $R = \dfrac{24 \times 24}{24+24} = \mathbf{9\ \Omega}$.

(C) 9 Ω

---

Points 1 and 2 are at diagonally opposite corners of the square. Current can travel two ways:

• Path 1 (A→B→2): two 12 Ω resistors in series = 24 Ω
• Path 2 (A→D→C→2 or equivalent): two 12 Ω resistors in series = 24 Ω

Both paths (24 Ω each) are in parallel between points 1 and 2:
$$R = \frac{24 \times 24}{24 + 24} = \frac{576}{48} = 9\ \Omega$$

The correct answer is (C) 9 Ω. (Option D, 6 Ω, is a common error from miscounting resistors in each path.)

(B) 110 W

P = VI = 220 V × 500 mA = 220 × 0.5 A = 110 W.

Use P = VI. Convert 500 mA to 0.5 A before multiplying. This matches Example 11.12 in the textbook exactly. Examiners expect the formula, substitution, and unit.

I notice that sub-question (c) appears to be incomplete — its actual question text is missing. I will answer (a) and (b) fully, and note the issue for (c).

---

(a) The live wire has a red insulation cover and the neutral wire has a black insulation cover. (The earth wire has a green insulation cover.)

(b)
$$I = \frac{P}{V} = \frac{1000 \text{ W}}{220 \text{ V}} \approx 4.5 \text{ A}$$

The current rating of the circuit should be at least 5 A.

(c) (The question text for sub-question (c) is missing/incomplete in the paper provided. Please supply the full question for a complete answer.)

Source: Chapter 12 – Electricity, Section: Domestic Electric Circuits

---

• (a) Examiners expect the standard wire colour codes: Live = Red, Neutral = Black, Earth = Green. Mentioning all three is safe even if only two are asked about.
• (b) Use P = VI → I = P/V. Always show the formula and substitution for full marks. Round up to the nearest standard fuse/circuit rating (5 A here).
• (c) Since the sub-question text is cut off, no answer can be given. Check your original paper and match the topic (likely about earthing, fuse, or short circuit) to standard textbook points.

(a) The resistance of a cylindrical conductor is given by:

$$R = \rho \frac{l}{A}$$

where $l$ = length, $A$ = area of cross-section, and $\rho$ = resistivity.

Rearranging: $\rho = \dfrac{RA}{l}$

SI unit of $\rho$ = $\dfrac{\Omega \times m^2}{m}$ = Ω m

(b) Alloys are used in electrical heating devices because:

• Their resistivity is higher than that of pure metals, so they produce more heat.
• They do not oxidise (burn) readily at high temperatures.

Source: Chapter 11, Section 11.5

---

• For part (a), always write the formula first, then rearrange to get ρ, then substitute SI units (Ω for R, m² for A, m for l) to derive Ω m. Examiners expect the derivation of the unit, not just stating it.
• For part (b), two points are needed but since it is 1 mark, one complete reason (high resistivity + no oxidation) suffices. Both points together make a stronger answer.

(Note: Since the exact circuit diagram is not visible, the standard textbook circuit assumed has 10 Ω and 15 Ω in parallel, combined with a 5 Ω series resistor, connected to a 6 V source.)

(a) Total Resistance:

Parallel combination of 10 Ω and 15 Ω:

$$\frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30}$$

$$R_p = 6 \text{ Ω}$$

Total resistance = 6 + 5 = 11 Ω (including 5 Ω series resistor)

(b) Total Current:

$$I = \frac{V}{R} = \frac{6}{6} = \mathbf{1 \text{ A}}$$

(Using only the parallel combination if no series resistor is specified, or per the given source value.)

(c) Potential difference across parallel combination:

$$V_p = I \times R_p = 1 \times 6 = \mathbf{6 \text{ V}}$$

---

• The exact numbers depend on the printed circuit diagram (source voltage and any series resistor). The method shown is universal: find R_p using the parallel formula, add series resistances for total R, apply V = IR for total current, then V_parallel = I × R_p.
• Examiners award marks for correct formula, substitution, and unit at each step — even if arithmetic differs due to the unseen diagram. Always show working clearly.
• For parallel resistors: use $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$; for series: $R_s = R_1 + R_2$.

Electric power is the rate at which electric energy is dissipated or consumed in an electric circuit. It is given by:
$$P = VI = I^2R = V^2/R$$

The SI unit of electric power is watt (W).

The power consumed in a circuit is said to be 1 watt when a current of 1 ampere flows through it at a potential difference of 1 volt.
$$1 \text{ W} = 1 \text{ V} \times 1 \text{ A}$$

Source: Chapter 11, Section 11.8

---

• The definition should mention "rate of consumption/dissipation of energy."
• For the second part, examiners expect the specific condition: 1 A current at 1 V potential difference. State the formula $1\text{ W} = 1\text{ VA}$ for full marks.
• Both marks are straightforward: 1 mark for definition + formula, 1 mark for the condition for 1 watt.

When wire of resistance R is cut into three equal parts, each part has resistance = R/3.

When these three parts are connected in parallel:

$$\frac{1}{R_p} = \frac{1}{R/3} + \frac{1}{R/3} + \frac{1}{R/3} = \frac{3}{R} + \frac{3}{R} + \frac{3}{R} = \frac{9}{R}$$

$$\therefore R_p = \frac{R}{9}$$

The total resistance of the combination is R/9.

Source: Chapter 11, Section 11.6.2 – Resistors in Parallel

---

• Step 1 (1 mark): Each part has resistance R/3 (resistance is proportional to length, so cutting into 3 equal parts gives R/3 each).
• Step 2 (1 mark): Apply the parallel formula $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ to get R/9.
• Examiners want both steps shown clearly. Simply writing R/9 without working will lose the method mark.

(Note: Since the actual circuit diagram values are not visible in the source, I will solve using the standard NCERT textbook example values: two resistors of 5 Ω and 10 Ω in parallel between P and Q, connected in series with a 3 Ω resistor, powered by a 6 V battery.)

(a) Total Resistance:

Parallel combination between P and Q:
$$\frac{1}{R_{PQ}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10} \Rightarrow R_{PQ} = \frac{10}{3} \ \Omega$$

Total resistance = $3 + \frac{10}{3} = \frac{19}{3} \approx 6.3 \ \Omega$

(b) Current from battery:

$$I = \frac{V}{R} = \frac{6}{\frac{19}{3}} = \frac{18}{19} \approx 0.95 \ \text{A}$$

(c) Potential difference between P and Q:

$$V_{PQ} = I \times R_{PQ} = \frac{18}{19} \times \frac{10}{3} \approx 3.16 \ \text{V}$$

Source: Chapter 11, Section 11.6 – Resistance of a System of Resistors

---

• The circuit diagram was not fully readable, so standard combination-circuit values were used. Always identify series and parallel parts first.
• For parallel resistors: use $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$; for series: add directly.
• Current from battery: apply Ohm's law $I = V/R_{total}$.
• Potential difference across any part: $V = I \times R_{that\ part}$.
• In board exams, show each formula, substitution, and unit clearly — each step carries marks.

(a)
(i) Battery (cell) — A battery maintains a constant potential difference in a circuit.
(ii) Rheostat (variable resistance) — A rheostat is used to change (vary) the electric current in a circuit.

(b)
Given: V₁ = 110 V, I₁ = 5 A

Resistance of heater: $R = \dfrac{V_1}{I_1} = \dfrac{110}{5} = 22\ \Omega$

When V₂ = 220 V:

$$I_2 = \frac{V_2}{R} = \frac{220}{22} = \textbf{10 A}$$

The current flowing through the heater will be 10 A.

Source: Chapter 11 – Electricity, Section 11.3 (Circuit components) and Example 11.4 (Ohm's law application)

---

• For (a)(i), a battery is the standard answer; some students write "cell" which is also acceptable.
• For (a)(ii), the answer expected is rheostat (variable resistance) — it is listed in Table 11.1 of the textbook.
• For (b), the key step is first finding resistance using Ohm's law (R = V/I), then applying it again at the new voltage. The resistance of the heater stays constant (metallic conductor). This mirrors Example 11.4 exactly.
• Always show formula, substitution, and unit — examiners award step marks.

(C) Ammeter in series and voltmeter in parallel

An ammeter measures current, so it must be in series to allow the same current to flow through it. A voltmeter measures potential difference, so it is connected in parallel across the component. This is clearly shown in Activities 11.4–11.6 of the textbook.

(C) Cases (ii) and (iii)

The bulb glows only when the circuit is complete and current flows through a conducting medium. Cases (ii) and (iii) involve conducting materials/solutions that allow current to flow, completing the circuit.

Since the diagram is not fully visible, option (C) is the standard answer for this classic NCERT-based activity question (Chapter 14, Acids Bases and Salts / Chapter 11 activity), where electrolyte solutions and metals conduct electricity, completing the circuit, while non-electrolytes (like sugar solution or distilled water) do not. Remember: conductors and electrolytes complete the circuit → bulb glows.

I notice sub-question (c) is incomplete — it mentions "Answer (i) or (ii)" but neither option is provided. I'll answer (a) and (b) fully, and note the gap for (c).

---

(a) The heat felt near the refrigerator is due to the heating effect of electric current. The compressor motor and condenser coils carry electric current; as current flows through the resistance of these components, electrical energy is converted into heat energy, which is released into the surroundings.

(b) By Ohm's law, $V = IR$.
Heat produced is given by Joule's law of heating:
$$H = VIt = I^2Rt = \frac{V^2}{R}t$$

(c) (Sub-question (c) appears incomplete — options (i) and (ii) were not provided. Please supply them for a complete answer.)

Source: Chapter — Effects of Electric Current (Electricity), NCERT Class 10 Science

---

• (a) Examiners expect you to name the phenomenon ("heating effect of electric current") and briefly explain that electrical energy converts to heat in the resistive parts of the refrigerator circuit.
• (b) You must show the derivation starting from $V = IR$ and arrive at all three equivalent forms of $H = I^2Rt$. Writing all three forms earns full credit.
• (c) Sub-question (c) is missing from the prompt. In the actual exam, it likely asks you to compare series vs. parallel heating, or calculate heat/power — answer whichever option (i)/(ii) is printed in your paper.

Graph: Plot V on the x-axis and I on the y-axis. The points (1.2, 0.4), (2.4, 0.8), (3.6, 1.2), (4.8, 1.6), (6.0, 2.0) lie on a straight line passing through the origin, showing that V ∝ I (Ohm's law is verified).

Calculation of Resistance:

Using Ohm's law: $R = \dfrac{V}{I}$

Taking any point, say V = 6.0 V, I = 2.0 A:

$$R = \frac{V}{I} = \frac{6.0 \text{ V}}{2.0 \text{ A}} = 3 \text{ Ω}$$

The resistance of the resistor is 3 Ω.

Source: Chapter 11, Section 11.4 — Ohm's Law

---

• Graph (1 mark): Plot the given points with V on x-axis and I on y-axis; the straight line through the origin confirms Ohm's law. CBSE expects you to mention the nature of the graph.
• Resistance (2 marks): Use $R = V/I$ with any consistent pair of values — all pairs give the same ratio (3 Ω) since V/I is constant. Show the formula, substitution, and unit.
• The slope of the V–I graph = V/I = R, so the graph itself gives resistance. Mentioning this earns full credit.

(a) When length is doubled, resistance doubles (R ∝ l). By Ohm's law, current halves.
∴ Ammeter reading = X/2 ampere

(b) When area of cross-section is doubled, resistance halves (R ∝ 1/A). Therefore, current doubles.
∴ Ammeter reading increases (becomes 2X ampere).

(c) Resistivity is defined as the resistance of a conductor of unit length and unit area of cross-section.
$$\rho = \frac{R \cdot A}{l}$$
SI unit: Ohm-metre (Ω·m)

Comparison: The resistivity of an alloy is generally higher than that of its constituent metals. For example, nichrome (an alloy) has much higher resistivity than nickel or chromium alone.

Source: Chapter 11 – Electricity, Resistance of a System of Resistors / Resistivity

---

• For (a) and (b), use R ∝ l and R ∝ 1/A directly, then apply V = IR (V constant from cell).
• For resistivity, examiners expect: definition, formula, SI unit, and the key comparison point (alloy > constituent metals). All four elements together fetch full 2 marks.
• Write "Ω·m" correctly — a common mistake is writing Ω/m.

Connection: Connect two resistors in parallel, then connect the third resistor in series with this parallel combination.

Diagram:

```
┌──[8Ω]──┐
A ──────┤ ├──────[8Ω]──── B
└──[8Ω]──┘
(parallel pair) (series)
```

Justification:

Equivalent resistance of the two resistors in parallel:

$$\frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} \implies R_p = 4 \ \Omega$$

Now this $R_p$ is connected in series with the third resistor:

$$R_{eq} = R_p + 8 = 4 + 8 = \mathbf{12 \ \Omega}$$

Hence the equivalent resistance of the combination is 12 Ω.

Source: Chapter 11, Section 11.6 – Resistance of a System of Resistors

---

• The key idea is to use a mixed (series-parallel) combination. Two resistors in parallel give half resistance (4 Ω); adding the third in series brings it back up to 12 Ω.
• Examiners expect: correct identification of the combination type, a clear circuit diagram, and step-by-step calculation using both formulas ($1/R_p$ for parallel and $R_s = R_1 + R_2$ for series).
• Always verify your final answer matches the required value.

When an electric heater is switched ON in a parallel circuit, it draws a large current due to its low resistance. This increases the total current drawn from the source, which causes a greater voltage drop across the internal resistance of the source. As a result, the terminal voltage (potential difference available to the circuit) decreases. Since the bulb now receives a lower voltage, the current through it decreases, and it glows dimmer (as power P = V²/R decreases).

Examiners expect two key points: (1) the heater draws large current, increasing drop across internal resistance, and (2) this reduces the voltage across the bulb, reducing its power/brightness. Mention of internal resistance is important — without it, voltage across parallel branches should stay constant, so the drop across internal resistance is the crux of the answer.

P — Ammeter (measures current in the circuit, connected in series)

Q — Nichrome wire / Resistor (the conducting wire whose resistance is being studied)

R — Plug key / Switch (used to open or close the circuit)

S — Cell / Battery (source of electric potential difference)

Source: Chapter 11 – Electricity, Section 11.5 (Activity 11.3, Fig. 11.5)

---

Since the actual diagram is not visible, the answer is based on Fig. 11.5 of Activity 11.3, which shows a standard circuit for studying resistance. Examiners expect students to name all four components correctly — name only (or name + one-line function) is sufficient for 2 marks (½ mark each). Always match the component to its role: ammeter in series, key to control circuit, cell as source, and the wire/resistor as the component under test.

(i) Let $n$ bulbs, each of resistance 8 Ω, be connected in parallel.

Equivalent resistance: $R_p = \dfrac{8}{n}$ Ω

Using Ohm's law: $I = \dfrac{V}{R_p}$

$$2 = \frac{4}{8/n} = \frac{4n}{8} = \frac{n}{2}$$

$$n = 4$$

4 bulbs should be connected in parallel.

---

(ii) An ammeter is used to measure electric current. It is always connected in series in the circuit, so that the same current flows through it.

---

(iii) Joule's Law of Heating: The heat produced in a resistor is

$$H = I^2 R t$$

It states that the heat produced in a conductor is directly proportional to (a) the square of the current ($I^2$), (b) the resistance ($R$) of the conductor, and (c) the time ($t$) for which current flows.

Source: Chapter 11 – Electricity, Sections 11.6.2 and 11.7

---

• Part (i): Use $R_p = R/n$ for $n$ identical resistors in parallel, then apply $V = IR_p$. Always show the formula and substitution.
• Part (ii): Many students confuse ammeter (series) with voltmeter (parallel). Examiner expects both the name and the connection method.
• Part (iii): State all three proportionalities clearly — examiners look for $I^2$, $R$, and $t$ explicitly. Writing just the formula without words may lose a mark.

(i) Resistance of a conductor is given by $R = \rho \dfrac{l}{A}$. When length or area of cross-section changes, the ratio $l/A$ changes, so R changes. However, resistivity ($\rho$) depends only on the nature (material) of the conductor, not on its dimensions. Hence, resistivity remains unchanged even when shape or size changes.

(ii) The resistivity of an alloy is generally higher than that of its constituent pure metals. Also, alloys do not oxidise (burn) readily at high temperatures. Therefore, they can withstand the high temperatures produced in heating devices like toasters and electric irons without melting or oxidising, making them more suitable than pure metals.

(iii) SI unit of electric current is ampere (A).
One ampere is defined as the flow of one coulomb of charge per second through a cross-section of a conductor.
$$1 \text{ A} = \frac{1 \text{ C}}{1 \text{ s}}$$

Source: Chapter 11 – Electricity, Sections 11.5 and 11.7.1

---

• For part (i), the key is distinguishing resistance (geometric + material property) from resistivity (material property only). Always quote the formula $R = \rho\,l/A$ to justify your reasoning.
• For part (ii), two reasons are expected: high resistivity (produces more heat efficiently) and non-oxidising nature at high temperatures. Both points carry marks.
• For part (iii), simply naming the unit alone may not fetch full marks. Always add the definition (1 C/s) when 1 mark is allocated to a "define" question.

Step 1: Find resistance of heater from rated values.

$$R = \frac{V}{I} = \frac{220}{11} = 20 \ \Omega$$

Step 2: Calculate power at 200 V using $P = V^2/R$.

$$P = \frac{V^2}{R} = \frac{(200)^2}{20} = \frac{40000}{20} = 2000 \ \text{W}$$

Power consumed = 2000 W

Source: Chapter 11, Section 11.8 Electric Power

---

• The heater's resistance is fixed (it's a resistive device), so first find R from rated voltage and current using Ohm's law.
• Since operating voltage changes, use $P = V^2/R$ (not $P = VI$ with rated current, because current will also change at the new voltage).
• Examiners award 1 mark for finding R correctly and 1 mark for the final power calculation.

Given: radius $r = 0.01$ cm $= 1 \times 10^{-4}$ m, length $l = 1.0$ cm $= 1 \times 10^{-2}$ m, $R = 7$ Ω

Area of cross-section: $A = \pi r^2 = 3.14 \times (10^{-4})^2 = 3.14 \times 10^{-8}$ m²

Using $R = \rho \dfrac{l}{A}$:

$$\rho = \frac{R \times A}{l} = \frac{7 \times 3.14 \times 10^{-8}}{1 \times 10^{-2}}$$

$$\rho = 21.98 \times 10^{-6} \approx 2.2 \times 10^{-5} \text{ Ω m}$$

Source: Chapter 11, Section 11.5

---

• The key formula is $\rho = RA/l$ (rearranged from $R = \rho l/A$).
• Always convert all units to SI (metres) before substituting.
• Examiners award 1 mark for correct substitution and 1 mark for the final answer with correct unit (Ω m). Don't forget to write the unit.

(i)
Given: V = 220 V, R = 55 Ω

Current, $I = \dfrac{V}{R} = \dfrac{220}{55} = \mathbf{4 \text{ A}}$

Power consumed, $P = VI = 220 \times 4 = \mathbf{880 \text{ W}}$

(ii)
Energy consumed by 3 bulbs in 30 days = 3 × 100 W × 5 h × 30 = 45,000 Wh = 45 kWh

Energy consumed by heater in 30 days = 1.0 kW × 0.5 h × 30 = 15 kWh

Total energy = 45 + 15 = 60 kWh

Cost = 60 × ₹3.60 = ₹ 216

(iii)
Alloys are used in heating devices because:

• The resistivity of alloys is higher than that of pure metals, so they produce more heat.
• Alloys do not oxidise (burn) readily at high temperatures, making them durable.

(Example: Nichrome, an alloy of Ni, Cr, Mn and Fe, is widely used.)

Source: Chapter 11 – Electricity, Sections 11.5 and 11.7

---

• (i) Direct application of Ohm's law ($I = V/R$) and power formula ($P = VI$). Both steps must be shown.
• (ii) Convert all power to kW and time to hours before multiplying. Add both energies, then multiply by rate. Show each step clearly.
• (iii) Examiners expect two reasons: high resistivity AND resistance to oxidation at high temperatures. These exact points are stated in Table 11.2's accompanying text in the textbook. Mentioning an example (Nichrome) adds value but is optional.

(i) Note: Since the actual circuit diagram values are not visible, a standard version commonly used in CBSE exams is assumed:
Battery = 6 V; Current through R1 = 1 A; Voltage across R1 = 2 V; Current through R2 = 0.5 A.

(I) Potential difference across R2:
V(R2) = Total voltage − Voltage across R1 = 6 − 2 = 4 V

(II) Value of R2:
Using Ohm's law: $R_2 = \dfrac{V_2}{I_2} = \dfrac{4}{0.5} = \mathbf{8\ \Omega}$

(III) Value of R1:
$R_1 = \dfrac{V_1}{I_1} = \dfrac{2}{1} = \mathbf{2\ \Omega}$

---

(ii) Factors on which resistance depends and derivation:

Resistance of a conductor depends on:

1. Length (l): Resistance increases with length.
2. Area of cross-section (A): Resistance decreases as area increases.
3. Nature of material (resistivity).

Derivation:

From experiments:
$$R \propto l \quad \text{and} \quad R \propto \frac{1}{A}$$

Combining:
$$R \propto \frac{l}{A}$$
$$\boxed{R = \rho \frac{l}{A}}$$

where $\rho$ (rho) is the electrical resistivity of the material (SI unit: Ω m), a characteristic property of the material.

Source: Chapter 11 – Electricity, Section 11.5

---

• For part (i), substitute the actual diagram values — the method (Ohm's law and Kirchhoff's voltage rule) stays the same.
• For part (ii), examiners expect: 3 factors named + both proportionality relations written + final formula with definition of ρ. Missing the definition of resistivity or its unit loses a mark.
• Write $R = \rho \frac{l}{A}$ clearly — this is the key derivation line.

(a) When key K is closed, current flows through the rod AB. Since AB carries current and is placed in a magnetic field, it experiences a force and gets displaced (moves/deflates upward or in a specific direction) due to the motor effect — the interaction between the magnetic field and the current-carrying conductor.

(b) Fleming's Left-Hand Rule determines the direction of force on the rod.

Statement: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the middle finger points in the direction of current, then the thumb points in the direction of force (motion) on the conductor.

---

• Part (a) is 1 mark: state what happens (rod moves/deflects) and why (force on current-carrying conductor in magnetic field).
• Part (b) is 2 marks: name the rule (½ mark) + state it correctly with all three fingers and what each represents (1½ marks). Many students lose marks by omitting the "mutually perpendicular" condition or mixing up which finger represents which quantity. Always mention all three: field → forefinger, current → middle finger, force/motion → thumb.

Given:
Charge on electron, $q = 1.6 \times 10^{-19}$ C
Potential difference, $V = 100$ V

Work done to move a charge across a potential difference is:

$$W = qV$$

$$W = 1.6 \times 10^{-19} \text{ C} \times 100 \text{ V}$$

$$W = 1.6 \times 10^{-17} \text{ J}$$

The work done required to move the electron is $1.6 \times 10^{-17}$ J.

---

• The key formula here is $W = qV$, which comes from the definition of potential difference: $V = W/Q$.
• The charge on one electron is always $1.6 \times 10^{-19}$ C — this must be remembered.
• The sign of charge (negative) is ignored when calculating magnitude of work done; examiners expect the positive value here.
• This is a standard 1-mark numerical; show the formula, substitution, and answer with unit.

One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

$$1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}}$$

Source: Chapter 11, Section 11.2 – Electric Potential and Potential Difference

---

• This is a standard 1-mark definition; one crisp sentence plus the formula is sufficient.
• Examiners expect the three key terms: 1 joule of work, 1 coulomb of charge, and two points in a conductor. Missing any one of these can cost marks.
• The formula $1\text{ V} = 1\text{ J C}^{-1}$ adds precision and is appreciated in board answers.

Current drawn by the heater:
$$I = \frac{P}{V} = \frac{2000 \text{ W}}{200 \text{ V}} = 10 \text{ A}$$

The heater requires 10 A, but the fuse is rated only 4 A. Since the current through the circuit (10 A) is much greater than the fuse rating (4 A), the fuse will blow immediately. Therefore, a 4 A fuse cannot be used with this heater.

Source: Chapter 11, Section 11.7.1

---

• Examiners expect you to calculate the current using $I = P/V$ — this is the key step that earns marks.
• Then compare the calculated current with the fuse rating and give a clear conclusion.
• The textbook example (1 kW iron at 220 V → 4.54 A → use 5 A fuse) follows the same logic; here 10 A >> 4 A, so the fuse is inadequate.
• Do not just state "no" without the calculation — that will lose a mark.