Answer:
52.2232 V
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 [tex]E=power\times time =75\times 60=4500J[/tex]
The energy storied in the capacitor is given by [tex]E=\frac{1}{2}CV^2[/tex]
[tex]4500=\frac{1}{2}3.3V^2[/tex]
[tex]V^2=2727.2727[/tex]
[tex]V=52.2232V[/tex]
