In the given capacitors circuit applied potential between $a$ and $b$ is $220\; V$. What is the equivalent capacitance of the network between $a$ and $b$? Given $C_1 = C_5 = 8.4 \mu F$ and $C_2 = C_3 = C_4 = 4.2\mu F$.

Given (required figure),
Potential applied $(V) = 220\;V$
Capacitors, $C_1 = C_5 = 8.4 \; \mu F$ and $C_2 = C_3 = C_4 = 4.2 \; \mu F$

Since, $C_3$ and $C_4$ are in series, there equivalent capacitance $C^{'}$ is,

$C^{'} = $ $ \frac{C_3 \; * \; C_4}{C_3 \; + \; C_4} = \frac{4.2 \;*\;4.2}{4.2\;+\;4.2}$ $ = 2.1 \; \mu F$

Now, $C^{'}$ and $C_2$ are parallel so,

$C^{' '} = C^{'} + C_2 = 2.1 \; + \; 4.2 = 6.3 \; \mu F$

Again,

$\frac{1}{C^{'''}} = \frac{1}{C_1} \; + \; \frac{1}{C_5} \; + \; \frac{1}{C^{''}} = \frac{1}{8.4} + \frac{1}{8.4} + \frac{1}{6.3}$

∴ $C^{'''} = 2.52 \; \mu F$

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