Science — CBSE Class 10 board question

(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

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• 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.