In the circuit shown,the charge on the 5, mu | vedclass

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(B) Let the potential of point $O$ be $V_{O} = 0 \, V$.Let the potential of the upper junction be $x$.The capacitors are connected to the batteries as

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$16.36$

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(B) Let the potential of point $O$ be $V_{O} = 0 \, V$.Let the potential of the upper junction be $x$.The capacitors are connected to the batteries as shown. The potential at the left plate of the $2\, \mu F$ capacitor is $6\, V$ and the potential at the right plate of the $4\, \mu F$ capacitor is $6\, V$.Using nodal analysis at the upper junction $x$:Sum of charges on the plates connected to junction $x$ is zero: $q_{1} + q_{2} + q_{3} = 0$.$2(x - 6) + 4(x - 6) + 5(x - 0) = 0$.$2x - 12 + 4x - 24 + 5x = 0$.$11x = 36$.$x = \frac{36}{11} \, V$.The charge on the $5\, \mu F$ capacitor is $q_{3} = C_{3} \cdot V_{3} = 5 \cdot (x - 0) = 5 \cdot \frac{36}{11} = \frac{180}{11} \, \mu C$.$q_{3} \approx 16.36 \, \mu C$.

In the circuit shown,the charge on the $5\, \mu F$ capacitor is $........\mu C$.

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