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Capacitors in Parallel

Capacitors in Parallel

This figure (a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance \({C}_{\text{p}}\), we first note that the voltage across each capacitor is \(V\), the same as that of the source, since they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source. The total charge \(Q\) is the sum of the individual charges:


Using the relationship \(Q=\text{CV}\), we see that the total charge is \(Q={C}_{\text{p}}V\), and the individual charges are \({Q}_{1}={C}_{1}V\), \({Q}_{2}={C}_{2}V\), and \({Q}_{3}={C}_{3}V\). Entering these into the previous equation gives


Canceling \(V\) from the equation, we obtain the equation for the total capacitance in parallel \({C}_{\text{p}}\):


Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “” indicates the expression is valid for any number of capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

\({C}_{\text{p}}=1\text{.}\text{000 µF}+5\text{.}\text{000 µF}+8\text{.}\text{000 µF}=\text{14}\text{.000 µF}.\)

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in this figure (b).

Total Capacitance in Parallel, \({C}_{\text{p}}\)

Total capacitance in parallel \({C}_{\text{p}}={C}_{1}+{C}_{2}+{C}_{3}+\text{.}\text{.}\text{.}\)

More complicated connections of capacitors can sometimes be combinations of series and parallel. (See this figure.) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

Example: A Mixture of Series and Parallel Capacitance

Find the total capacitance of the combination of capacitors shown in this figure. Assume the capacitances in this figure are known to three decimal places (\({C}_{1}=\text{1.000 µF}\), \({C}_{2}=\text{5.000 µF}\), and \({C}_{3}=\text{8.000 µF}\)), and round your answer to three decimal places.


To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors \({C}_{1}\) and \({C}_{2}\) are in series. Their combination, labeled \({C}_{\text{S}}\) in the figure, is in parallel with \({C}_{3}\).


Since \({C}_{1}\) and \({C}_{2}\) are in series, their total capacitance is given by \(\cfrac{1}{{C}_{\text{S}}}=\cfrac{1}{{C}_{1}}+\cfrac{1}{{C}_{2}}+\cfrac{1}{{C}_{3}}\). Entering their values into the equation gives


Inverting gives

\({C}_{\text{S}}=0\text{.}\text{833 µF}.\)

This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum

\(\begin{array}{lll}{C}_{\text{tot}}& =& {C}_{\text{S}}+{C}_{\text{S}}\\ & =& 0\text{.}\text{833}\phantom{\rule{0.25em}{0ex}}\text{μF}+8\text{.}\text{000}\phantom{\rule{0.25em}{0ex}}\text{μF}\\ & =& 8\text{.}\text{833}\phantom{\rule{0.25em}{0ex}}\text{μF}.\end{array}\)


This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations of capacitors.


  • Total capacitance in series \(\cfrac{1}{{C}_{\text{S}}}=\cfrac{1}{{C}_{1}}+\cfrac{1}{{C}_{2}}+\cfrac{1}{{C}_{3}}+\text{.}\text{.}\text{.}\)
  • Total capacitance in parallel \({C}_{\text{p}}={C}_{1}+{C}_{2}+{C}_{3}+\text{.}\text{.}\text{.}\)
  • If a circuit contains a combination of capacitors in series and parallel, identify series and parallel parts, compute their capacitances, and then find the total.

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