Question

What is the potential difference between the plates of a 3.3 F capacitor that store sufficient energy to operate a 75-W light bulb for one minute?

Answer 1

52.2232 V

Step-by-step explanation:

We have given the capacitance of the capacitor C = 3.3 F

Power P = 75 W

The time for which the bulb is operated t = 1 minute = 60 sec

Energy
`E=power* time =75* 60=4500J`

The energy storied in the capacitor is given by
`E=(1)/(2)CV^2`

`4500=(1)/(2)3.3V^2`

`V^2=2727.2727`

`V=52.2232V`