HC Verma Chapter 33 Problem 10

May 19, 2026

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Problem Statement

An electric device of resistance $R = 10\,\Omega$ is connected to a $40\,\text{V}$ supply for $t = 300\,\text{s}$. Find (a) the current, (b) the power, and (c) the heat produced.

Given Information

$R = 10\,\Omega$
$V = 40\,\text{V}$
$t = 300\,\text{s}$

Physical Concepts & Formulas

Joule heating: electrical energy converts to thermal energy at rate $P = V^2/R = I^2R$. Total heat $H = Pt$. This is the operating principle of all resistive heating devices.

$I = V/R$
$P = V^2/R$ — Joule heating
$H = Pt$ — total heat

Step-by-Step Solution

Step 1 — Current: $$I = V/R = 40/10 = 4.00\,\text{A}$$

Step 2 — Power: $$P = V^2/R = 40^2/10 = 160.0\,\text{W}$$

Step 3 — Heat: $$H = Pt = 160.0\times300 = 48000.0\,\text{J}$$

Worked Calculation

$$I = 4.00\,\text{A};\quad P = 160.0\,\text{W};\quad H = 48000\,\text{J}$$

Answer

$$\boxed{I = 4.00\,\text{A},\quad P = 160.0\,\text{W},\quad H = 48000\,\text{J}}$$

The 10Ω device at 40V draws 4.00A and generates 48000J of heat in 300s. In kWh: 0.013333 kWh — the basis for electricity billing.

Physical Interpretation