Q.10
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
Ohm’s Law Power Dissipation: Doubled Current Effect on Resistor Power
Ohm’s law relates voltage, current, and resistance in electric circuits, directly impacting power dissipation. When current doubles in a constant voltage setup with fixed resistance, power quadruples due to the P = I²R formula.
Correct Answer
The power dissipated across the resistor becomes 4P. This follows from the power equation P = I²R, where doubling current to 2I yields P’ = (2I)²R = 4I²R = 4P.
Original Setup
- Initial current I = V/R from Ohm’s law V = IR.
- Original power P = VI = I²R = V²/R.
Effect of Doubling Current
New current 2I gives new power P’ = (2I) × V, but since voltage stays constant, use P’ = (2I)²R = 4P. The quadratic dependence on current causes this fourfold increase.
Option Explanations
| Option | Explanation |
|---|---|
| (A) P/2 | Incorrect; halving power would occur if current halved, as P ∝ I². |
| (B) P | Incorrect; unchanged power implies constant current, not doubled. |
| (C) 2P | Incorrect; doubling would happen if power depended linearly on current (P = VI), but ignores I²R relation. |
| (D) 4P | Correct, as derived from P = I²R. |
Power Formula Summary
| Formula | Original Power | Doubled Current Power |
|---|---|---|
| P = I²R | I²R = P | (2I)²R = 4P |
| P = V²/R | Constant (fixed V, R) | N/A (I doubling implies scenario adjustment) |
1 Comment
Vanshika Sharma
December 25, 20254 times