Question

When a current of 2 A flows through a wire for 2.5 s, the amount of heat liberated is 20 J. The resistance of the wire is

• A. 4 \( \Omega \)
• B. 3 \( \Omega \)
• C. 1 \( \Omega \)
• D. 2 \( \Omega \)
• E. 5 \( \Omega \)

Hint:

To find the resistance using the heat liberated, use the formula \( H = I^2 R t \), where \( I \) is the current, \( R \) is the resistance, and \( t \) is the time.

Answer 1

The Correct Option is D

The heat \( H \) liberated by a current \( I \) flowing through a resistor \( R \) over a time \( t \) is given by Joule’s Law: \[ H = I^2 R t \] Where: - \( H = 20 \, \text{J} \) (the heat liberated) - \( I = 2 \, \text{A} \) (current) - \( t = 2.5 \, \text{s} \) (time) Substituting the given values into the equation: \[ 20 = (2)^2 R \times 2.5 \] \[ 20 = 4 R \times 2.5 \] \[ 20 = 10 R \] \[ R = \frac{20}{10} = 2 \, \Omega \] Thus, the resistance of the wire is \( 2 \, \Omega \). Hence, the correct answer is (D) 2 \( \Omega \).