Question

F the resistance remains constant and the voltage doubles, what effect will that have on the power? 

Answer 1

P = I^2R= V*I = V(V/R)= V^2/R. Substitute V with 2V and p=(2V)^2/R = 4V^2/R, so power quadruples which is E

Answer 2

Answer:

The power will increase by a factor of 4

E is correct.

Explanation:

Given that,

The power of the circuit in form of voltage

$P=\dfrac{V^2}{R}$

If the resistance remains constant then

$P\proto V^2$

using this equation we will compare P and V

$\dfrac{P_1}{P_2}=\dfrac{V_1^2}{V_2^2}$

Where,

$P_1=P$

$V_1=V$

$V_2=2V$

Now, we substitute the value into formula  and we get

$\dfrac{P}{P_2}=\dfrac{V}{(2V)^2}$

$\dfrac{P}{P_2}=\dfrac{1}{4}$

$P_2=4P$

Hence, The power will increase by a factor of 4