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
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 = [tex]\sqrt{(2U/C)}[/tex]
So to find energy U = Pt = 75.0 W × 60 sec
U = 4500 J
now V =[tex]\sqrt{2 x 4500j /4.6 F}[/tex]
V = 44.232 = 44.2 V
