Question

A resistor develops 800 J of thermal energy in 20 s on applying a potential difference of 20 V. Its resistance is

• A. 20 \(\Omega\)
• B. 10 \(\Omega\)
• C. 40 \(\Omega\)
• D. 0.5 \(\Omega\)

Hint:

Remember the three main formulas for electrical power: \(P = VI\), \(P = I^2R\), and \(P = V^2/R\). Choose the formula that directly uses the quantities given in the problem (here, V and E/t) to avoid unnecessary intermediate calculations (like finding the current I).

Answer 1

The Correct Option is B

Step 1: Concept Identification: This problem concerns the heating effect of electric current, or Joule heating, where electrical energy is converted to thermal energy when a potential difference is applied across a resistor. The objective is to determine the resistance given the energy, time, and voltage.

Step 2: Governing Equations: Electrical power \(P\) dissipated is the rate of energy conversion: \( P = \frac{E}{t} \). Power dissipated in a resistor can also be expressed as \( P = \frac{V^2}{R} \). These equations will be used to find \(R\).

Step 3: Calculation Details:
Given: \(E = 800 \, \text{J}\), \(t = 20 \, \text{s}\), \(V = 20 \, \text{V}\).

Procedure: First, calculate power: \( P = \frac{800 \, \text{J}}{20 \, \text{s}} = 40 \, \text{W} \). Then, solve for resistance using \( P = \frac{V^2}{R} \), which rearranges to \( R = \frac{V^2}{P} \). Substituting values: \( R = \frac{(20 \, \text{V})^2}{40 \, \text{W}} = \frac{400 \, \text{V}^2}{40 \, \text{W}} = 10 \, \Omega \).

Step 4: Conclusion: The resistance is 10 \(\Omega\).