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?

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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?

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Answer 1 · 2015-08-02T16:54:29

$5.7 V$ $5.7"V"$

Explanation:

farside.ph.utexas.edu

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

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

This can be rearranged to:

$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$ $C=(C_(1)C_(2))/(C_(1)+C_(2))$

$C = \frac{3.91 \times 10^{- 6} \times 7.41 \times 10^{- 6}}{(3.91 + 7.41) \times 10^{- 6}}$ $C=(3.91xx10^(-6)xx7.41xx10^(-6))/((3.91+7.41)xx10^(-6))$

$C = 2.56 \times 10^{- 6} F$ $C=2.56xx10^(-6)"F"$

$= 2.56 μ F$ $=2.56mu"F"$

The energy of a capacitor is given by:

$E = \frac{1}{2} C V^{2}$ $E=1/2CV^(2)$

$E = \frac{1}{2} \times 2.56 \times 10^{- 6} \times 12^{2}$ $E=1/2xx2.56xx10^(-6)xx12^(2)$

$E = 1.84 \times 10^{- 4} J$ $E=1.84xx10^(-4)"J"$

For capacitors in parallel:

farside.ph.utexas.edu

$C = C_{1} + C_{2}$ $C=C_1+C_2$

$C = 3.91 + 7.41 = 11.32 μ F$ $C= 3.91+7.41=11.32mu"F"$

$E = \frac{1}{2} C V^{2}$ $E=1/2CV^2$

$V^{2} = \frac{2 E}{C}$ $V^(2)=(2E)/(C)$

$= \frac{2 \times 1.84 \times 10^{- 4}}{11.32 \times 10^{- 6}}$ $=(2xx1.84xx10^(-4))/(11.32xx10^(-6))$

$V^{2} = 32.5$ $V^(2)=32.5$

$V = 5.7 V$ $V=5.7"V"$

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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?

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