Students will understand and apply the equations governing capacitors hooked up in series and parallel.

### Key Equations

\begin{align*} C_{\text{parallel}} &= C_1 + C_2 + C_3 + \dots && \text{[5] Capacitors in parallel add like resistors in series}\\ \frac{1}{C_{\text{series}}} &= \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}+ \dots && \text{[6] Capacitors in series add like resistors in parallel} \end{align*}

#### Example 1

Two capacitors, one of \begin{align*}10;\mu\text{F}\end{align*}

##### Solution

(a): To find the total capacitance, we'll use the equation give above for determining the equivalent capacitance of capacitors in series.

\begin{align*} \frac{1}{C_{total}}&=\frac{1}{C_1} + \frac{1}{C_2}\\ \frac{1}{C_{total}}&=\frac{1}{100\:\mu\text{F}} + \frac{1}{60\:\mu\text{F}}\\ C_{total}&=37.5\:\mu\text{F}\\ \end{align*}

(b): Since charge is the same across capacitors in series, we can use the charge found using the total capacitance and the total voltage drop to find the charge in the \begin{align*}C_1\end{align*}

\begin{align*} Q&=C_{total}V\\ Q&=37.5\:\mu\text{F} * 10\;\text{V}\\ Q&=375\:\mu\text{C}\\ \end{align*}

(c): Since we know the charge and the capacitance of \begin{align*}C_2\end{align*}

\begin{align*} Q&=C_2V_2\\ V_2&=\frac{Q}{C_2}\\ V_2&=\frac{375\:\mu\text{C}}{60\:\mu\text{F}}\\ V_2&=6.2\:\text{V}\\ \end{align*}

#### Example 2

The two capacitors used in the previous example problem are now connected to the battery in parallel. What is (a) the total capacitance and (b) the charge on \begin{align*}C_1\end{align*}

##### Solution

(a): To find the total capacitance, we'll us the equation given above for capacitors in parallel.

\begin{align*} C_{total}&=C_1+C_2\\ C_{total}&=100\:\mu\text{F} + 60\:\mu\text{F}\\ C_{total}&=160\:\mu\text{F}\\ \end{align*}

(b): Now, since the voltage across capacitors in parallel is equal, we can find the charge on \begin{align*}C_2\end{align*}.

\begin{align*} Q_2&=C_2V\\ Q_2&=60\:\mu\text{F} * 10\:\text{V}\\ Q_2&=600\:\mu\text{C}\\ \end{align*}

### Watch this Explanation

### Simulation

Note: go to the third tab to see circuits with multiple capacitors.

Capacitor Lab (PhET Simulation)

### Explore More

- You have two \begin{align*}42 \mu \mathrm{F}\end{align*} and one \begin{align*}39 \mu \mathrm{F}\end{align*} all wired in parallel. Draw the schematic and calculate the total capacitance of the system .
- You have two \begin{align*}42 \mu \mathrm{F}\end{align*} and one \begin{align*}39 \mu \mathrm{F}\end{align*} all wired in series. Draw the schematic and calculate the total capacitance of the system .
- Given a capacitor with \begin{align*}1\;\mathrm{cm}\end{align*} between the plates a field of \begin{align*}20,000\;\mathrm{N/C}\end{align*}is established between the plates.
- What is the voltage across the capacitor?
- If the charge on the plates is \begin{align*}1 \mu \mathrm{C}\end{align*}, what is the capacitance of the capacitor?
- If two identical capacitors of this capacitance are connected in series what it the total capacitance?
- Consider the capacitor connected in the following circuit at point \begin{align*}B\end{align*} with two switches \begin{align*}S\end{align*} and \begin{align*}T\end{align*}, a \begin{align*}20 \Omega\end{align*} resistor and a \begin{align*}120\;\mathrm{V}\end{align*}power source:
- Calculate the current through and the voltage across the resistor if \begin{align*}S\end{align*} is open and \begin{align*}T\end{align*} is closed
- Repeat if \begin{align*}S\end{align*} is closed and \begin{align*}T\end{align*} is open

**Answers**

- \begin{align*}123 \mu \mathrm{F}\end{align*}
- \begin{align*}0.073 \mu \mathrm{F}\end{align*}
- a. \begin{align*}6\mathrm{V}\end{align*} b. \begin{align*}0.3\mathrm{A}\end{align*} c. \begin{align*}18\mathrm{V}\end{align*} d. \begin{align*}3.6 \times 10^{-4}\mathrm{C}\end{align*} e. \begin{align*}3.2 \times 10^{-3}\mathrm{J}\end{align*} f. i) \begin{align*}80 \mu \mathrm{F}\end{align*} ii) \begin{align*}40\mu \mathrm{F}\end{align*} iii) \begin{align*}120 \mu \mathrm{F}\end{align*}