<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Capacitors in Series and Parallel

## A group of capacitors in series all have the same stored charge, a group of capacitors in parallel all have the same voltage.

Estimated6 minsto complete
%
Progress
Practice Capacitors in Series and Parallel

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated6 minsto complete
%
Capacitors in Series and Parallel

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.

Key Equations
Cparallel1Cseries=C1+C2+C3+=1C1+1C2+1C3+[5] Capacitors in parallel add like resistors in series[6] Capacitors in series add like resistors in parallel\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 10;μF\begin{align*}10;\mu\text{F}\end{align*} (C1\begin{align*}C_1\end{align*}) and one of 60μF\begin{align*}60\;\mu\text{F}\end{align*} (C2\begin{align*}C_2\end{align*}), are connected to a 10V battery in series. A diagram of the circuit is shown below. Determine (a) the total capacitance, (b) the charge stored on the 100μF\begin{align*}100\;\mu\text{F}\end{align*} capacitor, and (c) the voltage drop across the 60μF\begin{align*}60\;\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.

1Ctotal1CtotalCtotal=1C1+1C2=1100μF+160μF=37.5μF\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 C1\begin{align*}C_1\end{align*} capacitor.

QQQ=CtotalV=37.5μF10V=375μC\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 C2\begin{align*}C_2\end{align*}, we can find the voltage drop.

QV2V2V2=C2V2=QC2=375μC60μF=6.2V\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 C1\begin{align*}C_1\end{align*}. A diagram of the circuit is shown below.

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

CtotalCtotalCtotal=C1+C2=100μF+60μF=160μF\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 C2\begin{align*}C_2\end{align*}.

Q2Q2Q2=C2V=60μF10V=600μC\begin{align*} Q_2&=C_2V\\ Q_2&=60\:\mu\text{F} * 10\:\text{V}\\ Q_2&=600\:\mu\text{C}\\ \end{align*}

### Review

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

1. 123μF\begin{align*}123 \mu \mathrm{F}\end{align*}
2. 13.65μF\begin{align*}13.65 \mu \mathrm{F}\end{align*}
3. a. 200V\begin{align*}200\mathrm{V}\end{align*} b. 5x109F\begin{align*}5 x {10}^{-9}F\end{align*} c. 2.5x109F\begin{align*}2.5 x {10}^{-9}F\end{align*}  d. i. No current will flow and the voltage will be zero ii. V = 120V; I = 6A

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes