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Science — CBSE Class 10 board question

Q1. [3]
Answer the following:
  1. (i) State Joule's law of heating. Express it mathematically for the situation in which an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t.
  2. (ii) A 5 Ω resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in 10 s.
Previously asked in CBSE board exam
2022 31/1/1 Q12 (b)
Generated by claude-sonnet-4-6 · 2026-06-15 06:45 · grounding rag
Model Answer

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

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Explanation
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