Question

A 6 µF capacitor is first charged by connecting it to a battery (position 1). When the switch is moved to position 2, the capacitor is connected to an initially uncharged capacitor of capacitance 4 µF. What percentage of the stored energy in the 6 µF capacitor is dissipated when the switch is moved to position 2?

• A. 20%
• B. 40%
• C. 50%
• D. 60%

Answer 1

Answer: (B)

40%

Answer 2

Let the initial voltage of the battery be $V_B$. The 6 µF capacitor ($C_1$) is charged to this voltage. Initial charge on $C_1$: $Q_1 = C_1 V_B = 6 \mu F \cdot V_B$. Initial energy stored in $C_1$: $U_1 = \frac{1}{2} C_1 V_B^2 = \frac{1}{2} (6 \mu F) V_B^2$.

When the switch is moved to position 2, the 6 µF capacitor is connected in parallel with the uncharged 4 µF capacitor ($C_2$). The total charge $Q_{total} = Q_1 = 6 \mu F \cdot V_B$ is conserved. The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 6 \mu F + 4 \mu F = 10 \mu F$. The final common voltage across both capacitors is $V_f = \frac{Q_{total}}{C_{eq}} = \frac{6 \mu F \cdot V_B}{10 \mu F} = 0.6 V_B$.

The final energy stored in the system is $U_f = \frac{1}{2} C_{eq} V_f^2 = \frac{1}{2} (10 \mu F) (0.6 V_B)^2 = \frac{1}{2} (10 \mu F) (0.36 V_B^2) = \frac{1}{2} (3.6 \mu F) V_B^2$.

The energy dissipated is the difference between the initial energy of the 6 µF capacitor and the final energy of the system: $E_{dissipated} = U_1 - U_f = \frac{1}{2} (6 \mu F) V_B^2 - \frac{1}{2} (3.6 \mu F) V_B^2 = \frac{1}{2} (2.4 \mu F) V_B^2$.

The percentage of the stored energy in the 6 µF capacitor that is dissipated is: $\text{Percentage Dissipated} = \frac{E_{dissipated}}{U_1} \times 100\% = \frac{\frac{1}{2} (2.4 \mu F) V_B^2}{\frac{1}{2} (6 \mu F) V_B^2} \times 100\% = \frac{2.4}{6} \times 100\% = 0.4 \times 100\% = 40\%$