In the circuit shown in the figure, the potential difference across the $4.5\,\mu F$ capacitor is.......$volts$
Medium
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(d) The given circuit can be redrawn as follows potential difference across $4.5 \,\mu F$ capacitor
$V = \frac{9}{{\left( {\frac{9}{2} + 9} \right)}} \times 12$
$= 8\, V$
$V = \frac{9}{{\left( {\frac{9}{2} + 9} \right)}} \times 12$
$= 8\, V$
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