A 3.91 μF capacitor and a 7.41 μF capacitor are connected in series across a 12.0 V battery. What voltage would be required to charge a parallel combination of the same two capacitors to the same total energy?

1 Answer
Aug 2, 2015

5.7"V"

Explanation:

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The capacitance C of 2 capacitors in series like this is given by:

1/C=1/C_(1)+1/C_2

This can be rearranged to:

C=(C_(1)C_(2))/(C_(1)+C_(2))

C=(3.91xx10^(-6)xx7.41xx10^(-6))/((3.91+7.41)xx10^(-6))

C=2.56xx10^(-6)"F"

=2.56mu"F"

The energy of a capacitor is given by:

E=1/2CV^(2)

E=1/2xx2.56xx10^(-6)xx12^(2)

E=1.84xx10^(-4)"J"

For capacitors in parallel:

farside.ph.utexas.edufarside.ph.utexas.edu

C=C_1+C_2

C= 3.91+7.41=11.32mu"F"

E=1/2CV^2

V^(2)=(2E)/(C)

=(2xx1.84xx10^(-4))/(11.32xx10^(-6))

V^(2)=32.5

V=5.7"V"