Four identical capacitors are connected as shown in diagram. When a battery of $6 V$ is connected between $A$ and $B$, the charge stored is found to be $1.5\, \mu C$. The value of ${C_1}$ is........$\mu F$
Medium
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(d) The capacitance across $A$ and $B$
$ = \frac{{{C_1}}}{2} + {C_1} + {C_1} = \frac{5}{2}{C_1}$
As $Q = CV$,
$1.5\,\mu C = \frac{5}{2}{C_1} \times 6$
$==>$ ${C_1} = \frac{{1.5}}{{15}} \times {10^{ - 6}}$ $ = 0.1 \times {10^{ - 6}}F = 0.1\,\mu F.$
$ = \frac{{{C_1}}}{2} + {C_1} + {C_1} = \frac{5}{2}{C_1}$
As $Q = CV$,
$1.5\,\mu C = \frac{5}{2}{C_1} \times 6$
$==>$ ${C_1} = \frac{{1.5}}{{15}} \times {10^{ - 6}}$ $ = 0.1 \times {10^{ - 6}}F = 0.1\,\mu F.$
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