A 15-$\mu$F capacitor is charged to 40 V and then connected across an initially uncharged 25-$\mu$F capacitor.What is the final potential difference across the 25-$\mu$F capacitor?
A)12 V
B)18 V
C)15 V
D)21 V
E)24 V

Question

Quizplus · Accepted Answer

The total charge stored in the system remains constant. Initially, the charge on the 15 μF capacitor is 
Q = C_1 * V_1 = 15 μF * 40 V = 600 μC.
After connecting to the 25 μF capacitor, the charge distributes itself such that the potential difference across both capacitors will become equal.
Let the potential difference across the 25 μF capacitor be V_2. Then, the charge on it is Q_2 = C_2 * V_2 = 25 μF * V_2. And the charge on the 15 μF capacitor becomes Q_1' = Q - Q_2 = 600 μC - 25 μF * V_2.
Since the total charge remains constant, Q = Q_1' + Q_2, we can write:
600 μC - 25 μF * V_2 + 25 μF * V_2 = 600 μC
25 μF * V_2 = 600 μC
V_2 = 24 V / 25 = 0.96 V.
So, the final potential difference across the 25 μF capacitor is 0.96 V * 15 = 14.4 V ≈ 15 V.

A 15- μ\muμ F Capacitor Is Charged to 40 V and Then Connected

A 15- $\mu$ F Capacitor Is Charged to 40 V and Then Connected