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Wize University Physics Textbook (Master) > DC Circuits

Capacitors in Series and Parallel

Energy Stored in a CapacitorPrevious Section
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Capacitors in Series and Parallel

Capacitors in Series

Capacitors are said to be in series if they are wired one after the other along the same piece of wire.





The reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances:

 1Ceq=1C1+1C2+...+1Cn \boxed{\ \dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...+\dfrac{1}{C_n} \ }


The charge is the same for all capacitors:
 Q1=Q2=...=Qn \boxed{ \ Q_1=Q_2=...=Q_n \ }


The voltage drops across each capacitor add up to the total voltage:

 Vsource=V1+V2+...+Vn \boxed{\ V_{source}=V_1+V_2+...+V_n \ }





NOTE: You might see various notations used for the equivalent capacitance: it can also be called net, effective, total etc.

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Capacitors in Parallel

Capacitors are said to be in parallel if they are wired as separate branches, so that the charge splits up and recombines at the junctions.





The equivalent capacitance is the sum of the individual capacitances:


 Ceq=C1+C2+...+Cn \boxed{\ C_{eq}=C_1+C_2+...+C_n \ }


The voltage drops are the same for all capacitors:
 V1=V2=...=Vn \boxed{ \ V_1=V_2=...=V_n \ }


The charges on each capacitor add up to the total charge:

 Qtotal=Q1+Q2+...+Qn \boxed{\ Q_{total}=Q_1+Q_2+...+Q_n \ }




Wize Concept
  • The rules for combining resistors/capacitors are switched for series/parallel setups.
  • The charge for capacitors behaves like the current for resistors.


Exam Tip
You might see a combination of series and parallel capacitors in a circuit. If possible, put them together one by one to find the equivalent capacitance of the whole circuit.


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Example: Capacitors in Series and Parallel


Find the equivalent capacitance and the net charge stored in the capacitors for the circuit below, if all capacitors are identical (with capacitance CC). How much charge is drawn from the battery in terms of CC and VV?



First we'll combine capacitors 2 and 3, which are in series:

1C23=1C2+1C3=C2+C3C2C3\dfrac{1}{C_{23}}=\dfrac{1}{C_2}+\dfrac{1}{C_3}=\dfrac{C_2+C_3}{C_2C_3}

Then this combination is in parallel with capacitor 1:

C123=C23+C1C_{123}=C_{23}+C_1

Finally, this entire combination is in series with capacitor 4:

1Ceq=1C123+1C4\dfrac{1}{C_{eq}}=\dfrac{1}{C_{123}}+\dfrac{1}{C_4}


Let's see what happens when all capacitors are identical (and have capacitance CC ).

  • First combination:

1C23=1C2+1C3\dfrac{1}{C_{23}}=\dfrac{1}{C_2}+\dfrac{1}{C_3}

=1C+1C=\dfrac{1}{C}+\dfrac{1}{C}

=2C=\dfrac{2}{C}

This gives us C23=12 CC_{23}=\dfrac{1}{2} \ C .


  • Second combination:

C123=C23+C1C_{123}=C_{23}+C_1

=12 C+C=\dfrac{1}{2}\ C+C

=32 C=\dfrac{3}{2} \ C

So we have C123=32 CC_{123}=\dfrac{3}{2} \ C .

  • Third combination:

1Ceq=1C123+1C4\dfrac{1}{C_{eq}}=\dfrac{1}{C_{123}}+\dfrac{1}{C_4}

=132 C+1C=\dfrac{1}{\dfrac{3}{2} \ C}+\dfrac{1}{C}

=23C+1C=\dfrac{2}{3C}+\dfrac{1}{C}

=53 C=\dfrac{5}{3} \ C


Now for the charge drawn from the battery we have:

Qtotal=CeqVQ_{total}=C_{eq} V

=35 CV=\dfrac{3}{5} \ CV


checklist
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Practice: Circuit with a Switch


The switch (S) in the circuit below is connected to point A for a long time. What happens when we switch it to point B? Find the charge and voltage across each of the capacitors before and after we change the position of the switch from A to B.