A 3.30- mu F capacitor is charged by a 12.0 V battery. It is disconnected from t

Question

A 3.30- mu F capacitor is charged by a 12.0 V battery. It is disconnected from the battery and then connected to an uncharged 4.50-mu F capacitor (see (Figure 1)) Determine the total stored energy before the two capacitors are connected Express your answer using three significant figures and include the appropriate units. Determine the total stored energy after they are connected. Express your answer using three significant figures and include the appropriate units. What is the change in energy? Express your answer using three significant figures and include the appropriate units.

Explanation / Answer

(A)
C1 = 3.30 uF
V = 12.0 V
Q = C1 * V = 3.33 * 12.0 uC
Q = 39.96 uC

Initial Stored Energy, P.Ein = 1/2 * C1*V^2
P.Ein = 1/2 * 3.30 * 10^-6 * 12.0^2 J
P.Ein = 2.376 * 10^-4 J

When they are connected, Total Charge will be re distributed among the two Capacitors !! Total Voltage will be same across them !!

q1 = C1 * V
q2 = C2 * V

q1 + q2 = Q
(C1 + C2) * V = Q
V = 39.96/(3.3 + 4.5)
V = 5.12 V

(B)
Final Stored Energy, P.Efin = 1/2 * C1*V^2 + 1/2 * C2 * V^2
P.Ein = 1/2 * 3.30 * 10^-6 * 5.12^2 + 1/2 * 4.5 * 10^-6 * 5.12^2
P.Ein = 1.022 * 10^-4 J

(C)
Change in Energy, P.E = 2.376 - 1.022 * 10^-4 J
Change in Energy, P.E = 1.354 * 10^-4 J

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