C2 = 20.0 μF, and C3 = 25.0 μF. If no capacitor can withstand a potential difference of more than 100 V without failure, what are (a) the magnitude of the maximum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?

Part A. The maximum potential difference between points A and B is 300 V.

Part B. The maximum energy stored between points A and B is 26300 Joules.

Part A.

Given that the capacitors are connected in the series combination. The voltage across the circuit is 100V. So the total voltage is the sum of the voltage across each capacitor.

Hence we can conclude that the maximum potential difference between points A and B is 300 V.

Part B.

The maximum capacitance across the circuit is calculated given below.

So the energy stored in the three capacitors is given below.

Hence we can conclude that the maximum energy stored between points A and B is 26300 Joules.

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A) 300 V

B) 44,447J

Explanation:

A) from the question, the capacitors are in series.

Therefore, the total potential between points A and B would be the sum total of potentials across each of the capacitors.

So, Maximum potential across A and B = [100 + 100 + 100]V = 300V

B) Let Cs be the effective capacitance of the series combination;

So, 1/Cs = (1/10) + (1/20) +(1/25)

1/Cs = 0.1 + 0.05 + 0.04

1/Cs = 0.19

Cs = 1/0.19 = 5.26 microF

Energy stored = (1/2) Cs(V)^(2)

= (1/2) x 5.26 x (130^(2)) = 44,447J

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