The charge accumulated on the capacitor connected in the following circuit is $\mu C($ Given $C=150 \mu F)$
$V _A+\frac{10}{3}(1)-6(1)=V _B$
$V _A-V _B=6-\frac{10}{3}=\frac{8}{3}$ volt
$Q=C\left(V _A-V _B\right)$
$=150 \times \frac{8}{3}=400 \mu C$
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