HC Verma Chapter 33 Problem 13

Problem Statement

An electric device of resistance $R = 40\,\Omega$ is connected to a $160\,\text{V}$ supply for $t = 120\,\text{s}$. Find (a) the current, (b) the power, and (c) the heat produced.

Given Information

$R = 40\,\Omega$
$V = 160\,\text{V}$
$t = 120\,\text{s}$

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: $$I = V/R = 160/40 = 4.00\,\text{A}$$

Step 2 — Power: $$P = V^2/R = 160^2/40 = 640.0\,\text{W}$$

Step 3 — Heat: $$H = Pt = 640.0\times120 = 76800.0\,\text{J}$$

Worked Calculation

$$I = 4.00\,\text{A};\quad P = 640.0\,\text{W};\quad H = 76800\,\text{J}$$

Answer

$$\boxed{I = 4.00\,\text{A},\quad P = 640.0\,\text{W},\quad H = 76800\,\text{J}}$$

The 40Ω device at 160V draws 4.00A and generates 76800J of heat in 120s. In kWh: 0.021333 kWh — the basis for electricity billing.

Physical Interpretation