We charge a capacitor of capacitance $C_{1} = 8.0 \, \mu F$ by connecting it to a source of potential difference $V_{0} = 120 \, V$. Once C_{1} is charged, the potential difference is disconnected.
- What is the charge $Q_{0}$ on $C_{1}$?
- What is the energy stored in $C_{1}$
- A second uncharged capacitor of capacitance $C_{2} = 4.0 \,\mu F$ is connected in parallel to $C_{1}$. What is the potential difference across each capacitor, and what is the charge on each capacitor? What is the total energy of the system?
1.
$$\begin{aligned} Q_{0} &= \frac{V}{C} \\ &= \frac{120}{8.0 \mu} \\ &= 960 \, \mu F\end{aligned}$$
2.
$$\begin{aligned} U &= \frac{1}{2} Q_{0} V \\ &= \frac{1}{2} (960 \mu) (120) \\ &= 0.058 \, J \end{aligned}$$
3.
The equivalent capacitance is:
$$\begin{aligned} C_{eq} &= C_{1} + C_{2} \\ &= 12.0 \, \mu F \end{aligned}$$
Since charge must be conserved,
$$\begin{aligned} V &= \frac{Q_{0}}{C_{eq}} \\ &= 80 \, V \end{aligned}$$
Charge on $C_{1}$:
$$\begin{aligned} Q_{1} &= C_{1} V \\ &= 640 \, \mu C \end{aligned}$$
Charge on $C_{2}$:
$$\begin{aligned} Q_{2} &= C_{2} V \\ &= 320 \, \mu C \end{aligned}$$
Total energy of system:
$$\begin{aligned} U &= \frac{1}{2} Q_{1} V + \frac{1}{2} Q_{2} V \\ &= \frac{1}{2} Q_{0} V \\ &= 0.038 \, J \end{aligned}$$
Notice that the total energy decreased.
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