Question

A 2.0 μF capacitor is charged by a 12V battery. It is disconnected from the battery and then connected to an uncharged 5.0 μF capacitor. Determine the total stored energy (a) before the two capacitors are connected and (b) after they are connected.

Answer 1

(a)

$W_1=\frac{C_1U_1^2}{2}=\frac{2\cdot10^{-6}\cdot 12^2}{2}=144\cdot 10^{-6}J$ Answer

(b)

If the capacitors are connected in parallel that

$C=C_1+C_2=2+5=7\mu F$

$q_1=C_1\cdot U_1$

$U=\frac{q_1}{C}=\frac{C_1U_1}{C}=\frac{2\cdot10^{-6}\cdot12}{7\cdot10^{-6}}=3.43V$

$W_2=\frac{CU^2}{2}=\frac{7\cdot10^{-6}\cdot3.43^2}{2}=41\cdot10^{-6}J$ Answer