1. In a high-voltage electrical circuit, the current I is increased by 25% and the internal resistance R is decreased by 11%. If the power dissipated is given by $P = I^2R$, what is the net percentage change in the power P?
A) 14% increase
B) 26.5625% increase
C) 39.0625% increase
D) 11.25% decrease
Correct Answer: C
Let initial current be I and resistance be R. Initial Power $P = I^2R$.<br>New current I' = 1.25I. New resistance R' = 0.89R.<br>New Power $P' = (1.25I)^2 \times (0.89R) = 1.5625I^2 \times 0.89R = 1.390625 I^2R$.<br>Percentage change = $(1.390625 - 1) \times 100 = 39.0625%$ increase.