HC Verma Chapter 33 Problem 4

Problem Statement

An electric device of resistance $R = 50\,\Omega$ is connected to a $200\,\text{V}$ supply for $t = 300\,\text{s}$. Find (a) the current, (b) the power, and (c) the heat produced.

Given Information

  • $R = 50\,\Omega$
  • $V = 200\,\text{V}$
  • $t = 300\,\text{s}$

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: $$I = V/R = 200/50 = 4.00\,\text{A}$$

Step 2 — Power: $$P = V^2/R = 200^2/50 = 800.0\,\text{W}$$

Step 3 — Heat: $$H = Pt = 800.0\times300 = 240000.0\,\text{J}$$

Worked Calculation

$$I = 4.00\,\text{A};\quad P = 800.0\,\text{W};\quad H = 240000\,\text{J}$$

Answer

$$\boxed{I = 4.00\,\text{A},\quad P = 800.0\,\text{W},\quad H = 240000\,\text{J}}$$

The 50Ω device at 200V draws 4.00A and generates 240000J of heat in 300s. In kWh: 0.066667 kWh — the basis for electricity billing.

Physical Interpretation

The 50Ω device at 200V draws 4.00A and generates 240000J of heat in 300s. In kWh: 0.066667 kWh — the basis for electricity billing.


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