Question

Three capacitors C1 = 10.4 µF, C2 = 20.0 µF, and C3 = 29.5 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum potential difference across the series combination.

Answer 1

Answer:

66.75 V

Explanation:

A capacitor is a device that stores electrical energy in an electric field. It is made up of two conductor plates separated by a dielectric. It stores electrical charges which then produces a potential difference between the plates.

The equivalent capacitance for capacitors in series is given as:

$\frac{1}{C_{eq}}=\frac{1}{C_1} +\frac{1}{C_2}+\frac{1}{C_3} \\ \\\frac{1}{C_{eq}}=\frac{1}{10.4} +\frac{1}{20}+\frac{1}{29.5}\\ \\C_{eq}=5.55\ \mu F$

$The\ total\ charge\ stored(Q)\ is:\\\\Q=C_{eq}V\\\\Q=5.55*10^{-6}F*125\ V=694 \mu C$

The potential difference on each capacitor is:

$V_1=\frac{Q}{C_1}=\frac{694\mu C}{10.4\mu F} =66.75\ V\\\\V_2=\frac{Q}{C_2}=\frac{694\mu C}{20\mu F} =34.71\ V\\\\V_3=\frac{Q}{C_3}=\frac{694\mu C}{29.5\mu F} =23.53\ V$

The maximum potential difference is 66.75 V across the small capacitor