The charge accumulated on the capacitor connected in the following circuit is________ $\mu \mathrm{C}$ (Given $\mathrm{C}=150\ \mu \mathrm{F}$ )
JEE MAIN 2024, Diffcult
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$ V_A+\frac{10}{3}(1)-6(1)=V_B $
$ V_A-V_B=6-\frac{10}{3}=\frac{8}{3} \text { volt } $
$ Q=C\left(V_A-V_B\right) $
$ =150 \times \frac{8}{3}=400 \mu C$
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