# 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?

Oct 5, 2015

${V}_{p} = {V}_{s} \sqrt{\frac{\sqrt{{C}_{1}^{2} + {C}_{2}^{2}}}{{C}_{1} + {C}_{2}}}$

#### Explanation:

The energy stored in a capacitor is:

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

$U = \frac{q V}{2}$

The capacitance of the parallel case can be calculated as:
${C}_{p} = {C}_{1} + {C}_{2}$

And the capacitance of the series case can be calculated as:
${C}_{s} = \sqrt{{C}_{1}^{2} + {C}_{2}^{2}}$

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

${U}_{s} = {U}_{p}$
$\frac{{C}_{s} {V}_{s}^{2}}{2} = \frac{{C}_{p} {V}_{p}^{2}}{2}$
$\frac{{V}_{s}^{2} \sqrt{{C}_{1}^{2} + {C}_{2}^{2}}}{{C}_{1} + {C}_{2}} = {V}_{p}^{2}$

$\sqrt{\frac{{V}_{s}^{2} \sqrt{{C}_{1}^{2} + {C}_{2}^{2}}}{{C}_{1} + {C}_{2}}} = {V}_{p}$

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