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