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
Oct 5, 2015

Answer:

#V_p = V_ssqrt((sqrt(C_1^2 + C_2^2))/(C_1 + C_2))#

Explanation:

The energy stored in a capacitor is:

#U = 1/2 q^2 / C#
Which can also be written in the forms:
#U = (CV^2)/2#

#U = (qV)/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#
#(C_sV_s^2)/2 = (C_pV_p^2)/2#
#(V_s^2sqrt(C_1^2 + C_2^2))/(C_1 + C_2) = V_p^2#

#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.