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# Capacitors in Series and Parallel

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Practice Capacitors in Series and Parallel
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Capacitors in Series and Parallel

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

### Key Equations

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

Guidance
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 $10;\mu\text{F}$ ( $C_1$ ) and one of $60\;\mu\text{F}$ ( $C_2$ ), 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 $100\;\mu\text{F}$ capacitor, and (c) the voltage drop across the $60\;\mu\text{F}$ .

##### Solution

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

$\frac{1}{C_{total}}&=\frac{1}{C_1} + \frac{1}{C_2}\\\frac{1}{C_{total}}&=\frac{1}{100\:\mu\text{F}} + \frac{1}{60\:\mu\text{F}}\\C_{total}&=37.5\:\mu\text{F}\\$

(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 $C_1$ capacitor.

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

(c): Since we know the charge and the capacitance of $C_2$ , we can find the voltage drop.

$Q&=C_2V_2\\V_2&=\frac{Q}{C_2}\\V_2&=\frac{375\:\mu\text{C}}{60\:\mu\text{F}}\\V_2&=6.2\:\text{V}\\$

#### 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 $C_1$ . A diagram of the circuit is shown below.

##### Solution

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

$C_{total}&=C_1+C_2\\C_{total}&=100\:\mu\text{F} + 60\:\mu\text{F}\\C_{total}&=160\:\mu\text{F}\\$

(b): Now, since the voltage across capacitors in parallel is equal, we can find the charge on $C_2$ .

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

### Simulation

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

### Time for Practice

1. You have two $42 \mu \mathrm{F}$ and one $39 \mu \mathrm{F}$ all wired in parallel. Draw the schematic and calculate the total capacitance of the system .
2. You have two $42 \mu \mathrm{F}$ and one $39 \mu \mathrm{F}$ all wired in series. Draw the schematic and calculate the total capacitance of the system .
3. Given a capacitor with $1\;\mathrm{cm}$ between the plates a field of $20,000\;\mathrm{N/C}$ is established between the plates.
1. What is the voltage across the capacitor?
2. If the charge on the plates is $1 \mu \mathrm{C}$ , 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 $B$ with two switches $S$ and $T$ , a $20 \Omega$ resistor and a $120\;\mathrm{V}$ power source:
1. Calculate the current through and the voltage across the resistor if $S$ is open and $T$ is closed
2. Repeat if $S$ is closed and $T$ is open

1. $123 \mu \mathrm{F}$
2. $0.073 \mu \mathrm{F}$
3. a. $6\mathrm{V}$ b. $0.3\mathrm{A}$ c. $18\mathrm{V}$ d. $3.6 \times 10^{-4}\mathrm{C}$ e. $3.2 \times 10^{-3}\mathrm{J}$ f. i) $80 \mu \mathrm{F}$ ii) $40\mu \mathrm{F}$ iii) $120 \mu \mathrm{F}$