Question

What is the potential difference between the plates of a 4.6-F capacitor that stores sufficient energy to operate a 75.0-W light bulb for one minute?
How do i get the answer to be 44.2 V?

Answer 1

Answer:

V = 44.2 V

Explanation:

We have time = 1 minute = 60 sec, C = 4.6 F, P = 75.0 W

Solution:

Electric Potential Energy U Formula in capacitor for the given data.

U = 1/2 CV²

V = $\sqrt{(2U/C)}$

So to find energy U = Pt = 75.0 W × 60 sec

U = 4500 J

now V =$\sqrt{2 x 4500j /4.6 F}$

V = 44.232 = 44.2 V

Answer 2

Answer:

44.2v

Explanation:

The energy stored in capacitor, which is equal to energy required to operate a 75W bulb for one minute (60s) is

Energy = Pt =(75W)(60s) = 4500J

Solving equation energy = 1/2CV^2 for V we get

V = √2(Energy)/C

V = √2(4500J)/4.6F = 44.2V