Combination Capacitor Circuits

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Question #dd1e3

For capacitors in parallel:

The total capacitance $C_{T O T}$ $C_(TOT$ is given by:

$C_{T O T} = C_{1} + C_{2}$ $C_(TOT)=C_1+C_2$

Since $C_{T O T} = \frac{Q}{V}$ $C_(TOT)=Q/V$ the total can be found from $Q = C_{T O T} V$ $Q=C_(TOT)V$ where V is the voltage drop across the capacitors.

For capacitors in series:

$\frac{1}{C_{T O T}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$ $1/C_(TOT)=1/C_1+1/C_2$

So $Q = C_{T O T} V$ $Q=C_(TOT)V$

Michael · 1 · Jan 4 2015
What are some examples of combinations of capacitors?

Series, parallel and combinations of series and parallel/

There are four examples of combinations in the diagram. The following points show how to calculate the total capacitance of each combination.
1. Series
The equivalent capacitance, C , of the combination is worked out as follows:

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}$ $1/C=1/C_1+1/C_2+1/C_3$
or
$C = \frac{1}{1 / C_{1} + 1 / C_{2} + 1 / C_{3}}$ $C=1/(1//C_1+1//C_2+1//C_3)$

Total capacitance decreases in series.

2. Parallel
$C = C_{1} + C_{2} + C_{3}$ $C=C_1+C_2+C_3$

Total capacitance increases in parallel.

3. "Parallel in series"

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2} + C_{3}}$ $1/C=1/C_1+1/(C_2+C_3)$

4. "Series in parallel"

$C = \frac{1}{1 / C_{1} + 1 / C_{2}} + C_{3}$ $C=1/(1//C_1+1//C_2)+C_3$

Example with numbers based on combination 4.

$C_{1} = 200 μ F$ $C_1=200muF$ , $C_{2} = 400 μ F$ $C_2=400muF$ , $C_{3} = 400 μ F$ $C_3=400 muF$

$C = \frac{1}{1 / (200 \cdot 10^{- 6}) + 1 / (400 \cdot 10^{- 6})} + (400 \cdot 10^{- 6})$ $C=1/(1//(200*10^(-6))+1//(400*10^(-6)))+(400*10^(-6))$

$\Rightarrow C = \frac{1}{7500} + 400 \cdot 10^{- 6} = 133 \cdot 10^{- 6} + 400 \cdot 10^{- 6}$ $⇒C=1/7500+400*10^-6=133*10^-6+400*10^-6$

$C = 533 μ F$ $C=533muF$

Daniel W. · 1 · Dec 14 2014

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Combination Capacitor Circuits

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