A part of circuit is given here, if $V_A-V_B = 4V$ then find charge on $4\,\mu F$ capacitor.....$\mu C$
Medium
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From the figure we can see that potential at $B$ is $2 \vee$.
Therefore, potential at $\mathrm{A}$ is :
$V_{a}=4+V_{b}$
$V_{a}=4+2=6 V$
Therefore, charge on $4 \mu F$ capacitor is given by :
$q=C \times\left(V_{a}-V_{b}\right)$
$q=4 \times(6-6)$
$q=0 \mu C$
Therefore, option (A) is correct.
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