Question

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

A.The power will remain the same.
B.The power will decrease by a factor of 2.
C.The power will decrease by a factor of 4.
D.The power will increase by a factor of 2.
E.The power will increase by a factor of 4

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

The power will increase by a factor of 4

E is correct.

Step-by-step explanation:

Given that,

The power of the circuit in form of voltage

`P=(V^2)/(R)`

If the resistance remains constant then

`P\proto V^2`

using this equation we will compare P and V

`(P_1)/(P_2)=(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

`(P)/(P_2)=(V)/((2V)^2)`

`(P)/(P_2)=(1)/(4)`

`P_2=4P`

Hence, The power will increase by a factor of 4