## Related questions with answers

For the capacitor network shown in Fig. we saw earlier, the potential difference across $a b$ is $48 \mathrm{~V}$. Find (a) the total charge stored in this network:

Solution

VerifiedEquivalent capacitance of capacitors in series:

$\begin{aligned} \frac{1}{{{C_{{\text{eq}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} + ...... \end{aligned}$

where:

$\begin{aligned} {C_{{\text{eq}}}} \Rightarrow {\text{Equivalent capacitance of capacitors in series}}, \\ {C_1},{C_2},{C_3} \to {\text{capacitance of each individual capacitor}}{\text{.}} \\ \end{aligned}$

When capacitors are connected in series, the reciprocal of the equivalent capacitance of a series combination equals the sum of the reciprocals of the individual capacitances. The magnitude of the charge is the same on all of the plates of all of the capacitors, but the potential differences across individual capacitors are, in general, different.

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