Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-µF capacitor when it is fully charged?

6.05*10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1*C2) / (C1+C2) ... Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25*50) / (25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV ... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67*10⁻⁶ F

Substitute into equation 2

Q = 33 (16.67*10⁻⁶)

Q = 5.5*10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C ... Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q = 5.5*10⁻⁴ C, C = 25 µF = 25*10⁻⁶ F

Substitute into equation 3

E = 1/2 (5.5*10⁻⁴) ² / 25*10⁻⁶

E = 6.05*10⁻³ J.