A 12 kV power supply is used to individually charge two supercapacitors, which i

Question

A 12 kV power supply is used to individually charge two supercapacitors, which initially are connected in parallel. The first capacitor (C1) has a capacitance of 10 F, and the second capacitor, C2, has a capacitance of 5.0 F. After both capacitors are fully charged, they are disconnected from the power supply and then directly connected to each other. Immediately after reconnected, C1 and C2 are in series with each other. After electrostatic equilibrium is reached, the charges are redistributed.

1) What is the voltage across each capacitor after capacitors are connected to each other and equilibrium is reached?

2) What is the final charge stored on C1?

Explanation / Answer

Q=CV

Capacitors in parallel have equal potential drops. Qeq = C1 + C2 (parallel)

q1 = C1V = 12x10^3x10x10^-6 = 0.12C

q2 = C2V = 12x10^3 x5x10^-6 = 0.05C

NOW, the voltage across each capacitor after capacitors are connected to each other and equilibrium is reached

is V1 = 12x10^3 x(5x10^-6/15x10^-6) [10 + 5 = 15 uF]

= 4 kv

V2 = 12x10^3 x(10x10^-6/15x10^-6) = 8 kv

Final charge stored in C1 is = 10x10^-6x4x10^3 = 0.04C

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