# A 7x10^-6F and 3x10^-6F capacitor are connected in series across a 24V battery? What voltage is required to charge a parallel combination of the two capacitors to the same total energy?

Jun 10, 2014

The voltage required is 11 V.

First calculate the total capacitance of each combination.

Series: ${C}_{S} = \frac{1}{\frac{1}{C} _ 1 + \frac{1}{C} _ 2} = \frac{1}{\frac{1}{7} + \frac{1}{3}} = 2.1 \mu F$

(Note that I did not convert the $\mu F$ into $F$ therefore my answer was in $\mu F$.)

Parallel: ${C}_{P} = {C}_{1} + {C}_{2} = 7 + 3 = 2.1 \mu F$

Next calculate the energy stored in the series arrangement:
E_S=1/2C_SV_S^2=0.5×2.1×10^-6×24^2=6.05×10^-4J

Now calculate the voltage that would give the parallel combination the same total energy stored (${E}_{P} = {E}_{S}$):

E_P=1/2C_PV_P^2 ⇒ V_P = sqrt((2E_P)/C_P) = sqrt((2×6.05×10^-4)/(10×10^-6))=11V