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In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
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8 volt
$\vec{A}+(\hat{\imath}-2 \hat{\jmath}+2 \hat{k})+(-2 \hat{i}+\hat{\jmath}-\hat{k})=\frac{\hat{\jmath}}{\begin{array}{l}\text { unit } \\ \text { vector } \\ \text { along } \\ y-a x i s\end{array}}$
$\vec{A}=\hat{i}+2 \hat{\jmath}-\hat{k}$
$|\vec{A}|=\sqrt{6}$
$\vec{A}=\hat{i}+2 \hat{\jmath}-\hat{k}$
$|\vec{A}|=\sqrt{6}$
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