Three capacitors C1 = 10.1 µF,

C2 = 23.0 µF,

and

C3 = 29.4 µF

are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Question

Marisol Austin

Physics

26 November, 13:29

Three capacitors C1 = 10.1 µF,

C2 = 23.0 µF,

and

C3 = 29.4 µF

are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

+1

Answers (1)

Know the Answer?

Charles · Answer 1

energy is 0.14 J

Explanation:

given data

capacitor C1 = 10.1 µF

capacitor C2 = 23.0 µF

capacitor C3 = 29.4 µF

charged = 125 V

to find out

maximum energy stored

solution

we will apply here formula that is

energy = 1/2 Q²/c ... 1

here c = 1 / (1/c1 + 1/c2 + 1/c3)

c = 1 / (1/10.1 + 1/23 + 1/29.4)

c = 5.66 µF

and Q = c1v1

Q = 10.1 * 125

Q = 1262.5 µC

so from equation 1

energy = 1/2 * 1262.5² / 5.66 = 140804.439

so energy is 0.14 J

Three capacitors C1 = 10.1 µF,C2 = 23.0 µF,andC3 = 29.4 µFare connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Three capacitors C1 = 10.1 µF,

C2 = 23.0 µF,

and

C3 = 29.4 µF

are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.