Three capacitors with capacitance C1 350 nF C2 250 nF and

Three capacitors with capacitance C1 = 3.50 nF, C2 = 2.50 nF, and C3 = 5.50 nF are wired to a battery with V= 16.0 V, as shown in the figure. What is the potential drop across capcitor C?

Solution

C₂ parallel with C₃

C_p = C₂ + C₃ = 2.50 + 5.50 = 8.0 nF

1/C_equ = 1/8.0 nF + 1/3.50

C_equ = 2.43 nF

Q = C * V = 2.43 nF * 16 = 38.9 nC

The potential drop across capacitor C₂

V₂ = Q / C_p = 38.9 nC / 8 nF

= 4.86 volts