Question

A 2.50- mu F capacitor is charged to 747 V and 6.80-mu F capacitor is charged to 564 V capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other What will to the potential difference across each and the charge on each? Determine the potential difference across the first capacitor. Determine potential difference across the second capacitor.

Answer 1

here,

capaciatnce 1 , C1 = 2.5 * 10^-6 F and V1 = 747 V

capacitance 2 , C2 = 6.8 * 10^-6 F and V2 = 564 V

charge on 1 , Q1 = C1 * V1

Q1= 1.8675 * 10^-3 C

Q2 = C2 * V2 = 3.8352 * 10^-3 C

A)

the charge across each capacitor , Q = (Q1 + Q2)/2

Q = 2.85 * 10^-3 C

the potential difference across 1, V1 = Q/C1

V1 = 1141 V

B)

the potential difference across 2 , V2 = Q/C2

V2 = 419 V