In the circuit shown, charge on the $5\, \mu F$ capacitor is$........\mu C$
JEE MAIN 2020, Diffcult
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Let potential of point $O \quad v _{0}=0$

Now, using junction analysis

We can say, $\quad q_{1}+q_{2}+q_{3}=0$

$2(x-6)+4(x-6)+5(x)=0$

$x=\frac{36}{11} \quad q_{3}=\frac{36(5)}{11}=\frac{180}{11}$

$q _{3}=16.36 \mu C$

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