A DC circuit has a single battery of voltage V = 10.0 V attached to a resistor R = 100 Ohms. If the voltage is increased to 20.0 V, then by what factor does the power emitted by the resistor increase?

Answer 1

P2/P1 = ((2V)^2/R)/(V^2/R)=
(4V^2/R)*R/V^2=4V^2/V^2 = 4

Alternate Method:

P1 = V1^2/R = 10^2/100 = 1. Watt.

P2 = V2^2/R = 20^2/100 = 4 Watts.

P2/P1 = 4W/1W = 4

So the power is increased by a factor of
4.