A 15-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 4.3 µF, 12.6 µF, and 31.2 µF. Find the voltage across the 31.2-µF capacitor.

1.394 V

Explanation:

When a capacitor is connected in series, The sum of a capacitance is given as

1/Ct = 1/C1 + 1/C2 + 1/C3 ... Equation 1

Where Ct = Value of the combined capacitor, C1 = first capacitance, C2 = Second capacitance, C3 = Third capacitance.

Given: C1 = 4.3 µF, C2 = 12.6 µF, C3 = 31.2 µF

Substitute into equation 1

1/Ct = 1/4.3 + 1/12.6 + 1/31.2

1/Ct = 0.233 + 0.0794 + 0.0321

1/Ct = 0.3445

Ct = 1/0.3445

Ct = 2.9 µF.

But

Q = CtV ... Equation 2

Where

Q = Amount of charge, V = voltage, C = total capacitance

Given: V = 15 V, Ct = 2.9 µF

Substitute into equation 1

Q = 15 (v) * 2.9 (µF)

Q = 43.5 µC

The voltage across the 31.2 µF is

V₃ = Q/31.2 µF ... equation 3

Where V₃ = The voltage across the 31.2 µF capacitor.

Note: When capacitors are connected in series, the same quantity of charge flows through them.

V₃ = 43.5 (µC) / 31.2 (µF)

V₃ = 1.394 V.

Hence the voltage across the 31.2 µF = 1.394 V