Question

An electric circuit with a resistor of constant resistance ‘R’ is maintained at a constant voltage ‘V’. Based on Ohm’s law, if the current ‘I’ through the circuit is doubled, the power ‘P’ dissipated across the resistor is

• A. \(P/2\)
• B. \(P\)
• C. \(2P\)
• D. \(4P\)

Hint:

Always remember: Power \(P\) in a resistor is proportional to \(I^2\). Doubling current makes power 4 times.

Answer 1

The Correct Option is D

Step 1: Recall power formula.
\[ P = I^2 R \] Step 2: Consider doubling the current.
If the current becomes \(2I\), then power becomes: \[ P' = (2I)^2 R = 4I^2R \] Step 3: Compare with original power.
Original power = \(P = I^2R\). Thus, \[ P' = 4P \] Step 4: Clarification about constant voltage.
Since \(V = IR\), doubling \(I\) would imply \(V\) changes. But if \(V\) is held constant externally, then \(I\) cannot double. However, as per the problem statement assumption, when current doubles, the mathematical relation yields \(P' = 4P\). Final Answer: \[ \boxed{4P} \]