Question

A parallel combination of two capacitors, C_{1} and
C_{2} where C_{2}=2C_{1}, is connected to a
battery. If the charge stored on C_{1} is 8.68
×10^{-6} C and the total energy U stored in the combination
is 14.83 ×10^{-9} Joule, then the capacitance of
C_{2} is:

Answer #1

Total capacitance of the parallel combination of C1 and C2, Ceq = (C1 + C2) = 3*C1

Charge stored in C1, Q1 = C1 * V = 8.68 X 10-6 C ( if V = battery voltage)

Total energy stored in the combination, U = (0.5) * Ceq * V2 = 14.83 X 10-9 J

( U / Q1) = ( (0.5) * Ceq * V2 / ( C1 * V )) = ( (0.5) * 3*C1 * V2 / ( C1 * V )) = 1.5*V

(U/Q1) = (14.83 X 10-9 / 8.68 X 10-6 ) = 1.5 * V

V = 1.1390 X 10-3 Volts

C1 = ( Q1 / V) = (8.68 X 10-6 ) / (1.1390 X 10-3) = 7.62 mF

C2 = 2* C1 = 15.24 mF (Ans)

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