Capacitors in parallel have the same voltage, but different charge stored. Capacitors in series have the same charge stored, but different voltages. Remember that if a capacitor are hooked up to the battery they will have the same voltage as the battery. If the capacitor is unhooked from a battery and other capacitors are attached to it, then the voltage can change but the total amount of charge must remain constant. Charge conservation holds that charge can not be created or destroyed. When solving problems involving capacitor circuits, we use the equation for the charge on a capacitor much like we use Ohm's Law.

#### Example 1

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

(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*}

(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*}

### Simulation

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

Capacitor Lab (PhET Simulation)

### Review

- You have two \begin{align*}42 \mu \mathrm{F}\end{align*}
42μF and one \begin{align*}39 \mu \mathrm{F}\end{align*}39μF 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*}
42μF and one \begin{align*}39 \mu \mathrm{F}\end{align*}39μF 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*}
1cm between the plates a field of \begin{align*}20,000\;\mathrm{N/C}\end{align*}20,000N/C 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*}
1μC , 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*}
B with two switches \begin{align*}S\end{align*}S and \begin{align*}T\end{align*}T , a \begin{align*}20 \Omega\end{align*}20Ω resistor and a \begin{align*}120\;\mathrm{V}\end{align*}120V power source:- Calculate the current through and the voltage across the resistor if \begin{align*}S\end{align*}
S is open and \begin{align*}T\end{align*}T is closed - Repeat if \begin{align*}S\end{align*}
S is closed and \begin{align*}T\end{align*}T is open

- Calculate the current through and the voltage across the resistor if \begin{align*}S\end{align*}

**Review (Answers)**

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