A 3.96 μF capacitor and a 7.94 μF capacitor are connected in series across a 14.0 V battery. What voltage would be required to charge a parallel combination of the same two capacitors to the same total energy?

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A 3.96 μF capacitor and a 7.94 μF capacitor are connected in series across a 14.0 V battery. What voltage would be required to charge a parallel combination of the same two capacitors to the same total energy?

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Answer 1 · 2015-10-05T14:06:09

$V_{p} = V_{s} \sqrt{\frac{\sqrt{C_{1}^{2} + C_{2}^{2}}}{C_{1} + C_{2}}}$ $V_p = V_ssqrt((sqrt(C_1^2 + C_2^2))/(C_1 + C_2))$

Explanation:

The energy stored in a capacitor is:

$U = \frac{1}{2} \frac{q^{2}}{C}$ $U = 1/2 q^2 / C$
Which can also be written in the forms:
$U = \frac{C V^{2}}{2}$ $U = (CV^2)/2$

$U = \frac{q V}{2}$ $U = (qV)/2$

The capacitance of the parallel case can be calculated as:
$C_{p} = C_{1} + C_{2}$ $C_p = C_1 + C_2$

And the capacitance of the series case can be calculated as:
$C_{s} = \sqrt{C_{1}^{2} + C_{2}^{2}}$ $C_s = sqrt(C_1^2 + C_2^2)$

So we want to find a series energy $U_{s}$ $U_s$ that is the same as the parallel energy $U_{p}$ $U_p$ .

$U_{s} = U_{p}$ $U_s = U_p$
$\frac{C_{s} V_{s}^{2}}{2} = \frac{C_{p} V_{p}^{2}}{2}$ $(C_sV_s^2)/2 = (C_pV_p^2)/2$
$\frac{V_{s}^{2} \sqrt{C_{1}^{2} + C_{2}^{2}}}{C_{1} + C_{2}} = V_{p}^{2}$ $(V_s^2sqrt(C_1^2 + C_2^2))/(C_1 + C_2) = V_p^2$

$\sqrt{\frac{V_{s}^{2} \sqrt{C_{1}^{2} + C_{2}^{2}}}{C_{1} + C_{2}}} = V_{p}$ $sqrt((V_s^2sqrt(C_1^2 + C_2^2))/(C_1 + C_2)) = V_p$

I'll leave it as an exercise for the student to plug in the numbers required for this calculation.

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A 3.96 μF capacitor and a 7.94 μF capacitor are connected in series across a 14.0 V battery. What voltage would be required to charge a parallel combination of the same two capacitors to the same total energy?

1 Answer

Explanation:

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