Question

A parallel combination of a 1.17-μF capacitor and a 2.87-μF one is connected in series to a 4.47-μF capacitor. This three-capacitor combination is connected to a 17.5-V battery. Find the charge of each capacitor.

Answer 1

Equivalent capacitance = (1.17 + 2.87)*4.47/[1.17 + 2.87 + 4.47] = 2.12 uF

From Q = CV we get charge Q drawn from the battery = 2.12 x 10^-6 x 17.5 = 37.1 X 10^-6 C.

When capacitors are in series, the charge on each capacitor is same.

4.04 uF capacitor and 4.47 uF cap are in series. Hence, Q the charge on each of these is same (37.1 X 10^-6 C)

V = Q/C. Hence voltage across 4.47 uF cap = 37.1 X 10^-6 C/4.47x 10^-6 = 8.3 V

Voltage across 4.04 uF cap = 37.1 X 10^-6/4.04x 10^-6 = 9.18 V

As the voltage across capacitors in parallel is same, voltage across each of 1.17 & 2.87 uF caps will be 9.18 V.

So Charge on 1.17 uF = 1.17 uF*9.18 = 10.74 uC

and charge on 2.87 uF = 2.87 uF*9.18 = 26.35 uC