A capacitor,C₁ = 5.0 $\mu$F,is charged up to 8 V.It is then connected to a second uncharged capacitor C₂ = 2.5 $\mu$F.The voltage across the capacitor C₁ after the system has come to equilibrium is A)2.67 V B)4.0 V C)5.33 V D)8.0 V E)12.0 V

Question

Quizplus · Accepted Answer

In a system of capacitors connected in series, the charges on both capacitors are equal. Using the formula Q = CV, we can find the charge on the first capacitor: Q = 5F x 8V = 40C.

Since the charges on both capacitors are equal, the voltage on the second capacitor can be found using V = Q/C, where Q is the charge on the first capacitor and C is the capacitance of the second capacitor: V = 40C / 2.5F = 16V/3 = 5.33V (approx).

Therefore, the voltage across the first capacitor after equilibrium is 5.33V. Hence, the answer is choice C.

A Capacitor,C1 = 5 μ\muμ F,is Charged Up to 8 V μ\muμ

A Capacitor,C₁ = 5 $\mu$ F,is Charged Up to 8 V $\mu$