Capacitors and capacitance — practice questions

(a) In Series:

1. Use the series capacitance formula: $\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2}$ $\frac{1}{C_{total}} = \frac{1}{4.0 \times 10^{-6} \text{ F}} + \frac{1}{6.0 \times 10^{-6} \text{ F}}$ $\frac{1}{C_{total}} = (2.5 \times 10^5 + 1.667 \times 10^5) \text{ F}^{-1} = 4.167 \times 10^5 \text{ F}^{-1}$ $C_{total} = \frac{1}{4.167 \times 10^5} = 2.4 \times 10^{-6} \text{ F} = 2.4 \mu\text{F}$
2. Calculate total charge stored using $Q = C_{total}V$: $Q_{total} = (2.4 \times 10^{-6} \text{ F}) \times (12 \text{ V}) = 2.88 \times 10^{-5} \text{ C}$ Therefore, total capacitance is $2.4 \mu\text{F}$ and total charge is $28.8 \mu\text{C}$.

(b) In Parallel:

1. Use the parallel capacitance formula: $C_{total} = C_1 + C_2$ $C_{total} = 4.0 \mu\text{F} + 6.0 \mu\text{F} = 10.0 \mu\text{F}$
2. Calculate total charge stored using $Q = C_{total}V$: $Q_{total} = (10.0 \times 10^{-6} \text{ F}) \times (12 \text{ V}) = 1.2 \times 10^{-4} \text{ C}$ Therefore, total capacitance is $10.0 \mu\text{F}$ and total charge is $120 \mu\text{C}$.

(a) Initial Energy Stored: Use the formula $E = \frac{1}{2}CV^2$. $E_{initial} = \frac{1}{2} \times (2200 \times 10^{-6} \text{ F}) \times (10.0 \text{ V})^2$ $E_{initial} = 0.11 \text{ J}$

(b) Final Potential Difference:

1. First, find the initial charge stored on the first capacitor. This charge is conserved. $Q_{initial} = C_1 V_1 = (2200 \times 10^{-6} \text{ F}) \times (10.0 \text{ V}) = 0.022 \text{ C}$
2. When connected in parallel, this total charge is shared across the total capacitance. $C_{total} = C_1 + C_2 = 2200 \mu\text{F} + 4700 \mu\text{F} = 6900 \mu\text{F}$
3. The new potential difference is found using $V = Q/C$. $V_{final} = \frac{Q_{total}}{C_{total}} = \frac{0.022 \text{ C}}{6900 \times 10^{-6} \text{ F}} = 3.188... \text{ V} \approx 3.19 \text{ V}$

(c) Final Energy Stored: Use the total capacitance and final voltage. $E_{final} = \frac{1}{2} C_{total} V_{final}^2$ $E_{final} = \frac{1}{2} \times (6900 \times 10^{-6} \text{ F}) \times (3.188... \text{ V})^2$ $E_{final} = 0.0351... \text{ J} \approx 0.035 \text{ J}$

(d) Energy Lost: The energy lost is the difference between the initial and final stored energy. $E_{lost} = E_{initial} - E_{final} = 0.11 \text{ J} - 0.0351 \text{ J} = 0.0749 \text{ J} \approx 0.075 \text{ J}$ This energy is dissipated as heat in the connecting wires as the charge redistributes.