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In the given circuit, a charge of \(+80 \mu C\) is given to...

In the given circuit, a charge of \(+80 \mu C\) is given to the upper plate of the \(4 \mu F\) capacitor. Then in the steady state, the charge on the upper plate of the \(3 \mu F\) capacitor is

\(+32 \mu \mathrm{C}\)
\(+40 \mu \mathrm{C}\)
\(+48 \mu \mathrm{C}\)
\(+80 \mu \mathrm{C}\)
Solution:
The total charge on plate \(A\) will be \(80 \mu \mathrm{C}\).

\(2 \mu F\) and \(3 \mu F\) capacitors are in parallel. Therefore, \(C_{e q}=2+3=5 \mathrm{HF}\)

Charge on capacitor of \(3 \mu \mathrm{F}\) capacitance \(q=\frac{3}{5} \times 80=48 \mu C\)