College

Three capacitors with capacitances of 4 μF, 6 μF, and 8 μF are connected in series with a 13 V battery. What is the potential difference across the 6 μF capacitor?

Answer :

The potential difference across the 6 μF capacitor is V= 3.71 V.

How to solve

In a series circuit, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances.

Thus, [tex]C_{total[/tex] = 1/(1/4 + 1/6 + 1/8) μF = 1.714 μF.

Since they are in series, the same charge (Q) will store on each capacitor.

The charge stored on a capacitor is given by Q = CV.

Therefore, Q = 1.714 μF * 13 V = 22.282 μC.

The potential difference (V) across a capacitor is given by V = Q/C.

Therefore, the potential difference across the 6 μF capacitor is V = 22.282 μC / 6 μF = 3.71 V.

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