Question

A 3.0-mF capacitor and 5.0-mF capacitor are connected in a series across an 8.0-V potential source. What is the potential difference across the 3.0-mF capacitor?

A. 0 V
B. 2.7 V
C. 5.0 V
D. 3.6 V
E. 8.0 V

Answer 1

The potential difference across the 3.0-mF capacitor, when it is connected in series with a 5.0-mF capacitor across an 8.0-V potential source, is found to be 5.0 V after calculating the series equivalent capacitance and total charge.

Step-by-step explanation:

To find the potential difference across the 3.0-mF capacitor when it is connected in series with a 5.0-mF capacitor across an 8.0-V potential source, we first need to calculate the equivalent capacitance of the series combination. The equivalent capacitance (Ceq) of capacitors in series is determined using the formula:

1/Ceq = 1/C1 + 1/C2

Where C1 is the capacitance of the first capacitor (3.0 mF) and C2 is the capacitance of the second capacitor (5.0 mF). After calculating Ceq, we can find the total charge (Q) stored in the series combination using Q = Ceq * V where V is the potential source (8.0 V). The charge is the same for both capacitors in series, and using this charge, we can then determine the voltage across the 3.0-mF capacitor (V3.0mF) by using V3.0mF = Q/C3.0mF.

Carrying out the calculations:

1/Ceq = 1/(3.0 mF) + 1/(5.0 mF)

Ceq = 1.875 mF

Q = 1.875 mF * 8.0 V = 15.0 µC

V3.0mF = 15.0 µC / 3.0 mF

V3.0mF = 5.0 V

The potential difference across the 3.0-mF capacitor is therefore 5.0 V, which corresponds to option C.