Question:
A 10 \(\Omega\) resistor takes a current of 5 A. The thermal energy developed in 40 s is:
A 10 \(\Omega\) resistor takes a current of 5 A. The thermal energy developed in 40 s is:
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Always use \(H = I^2Rt\) for heating effect problems in resistors.
Updated On: Jun 20, 2026
- 5 kJ
- 15 kJ
- 7.5 kJ
- 10 kJ
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The Correct Option is D
Solution and Explanation
Step 1: Use Joule’s heating law.
Thermal energy produced in a resistor is: \[ H = I^2 R t \]
Step 2: Substitute given values.
\[ I = 5 \, A,\quad R = 10 \, \Omega,\quad t = 40 \, s \]
Step 3: Compute step by step.
\[ H = (5)^2 \times 10 \times 40 \] \[ H = 25 \times 10 \times 40 \] \[ H = 250 \times 40 = 10000 \, J \]
Step 4: Convert into kJ.
\[ 10000 \, J = 10 \, kJ \]
Step 5: Final interpretation.
Electrical energy fully converts into heat energy in the resistor.
\[ \boxed{10 \, kJ} \]
Thermal energy produced in a resistor is: \[ H = I^2 R t \]
Step 2: Substitute given values.
\[ I = 5 \, A,\quad R = 10 \, \Omega,\quad t = 40 \, s \]
Step 3: Compute step by step.
\[ H = (5)^2 \times 10 \times 40 \] \[ H = 25 \times 10 \times 40 \] \[ H = 250 \times 40 = 10000 \, J \]
Step 4: Convert into kJ.
\[ 10000 \, J = 10 \, kJ \]
Step 5: Final interpretation.
Electrical energy fully converts into heat energy in the resistor.
\[ \boxed{10 \, kJ} \]
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