JEE Advanced · Physics · 18. Capacitance
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
- A \(+32 \mu \mathrm{C}\)
- B \(+40 \mu \mathrm{C}\)
- C \(+48 \mu \mathrm{C}\)
- D \(+80 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(+48 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
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 \(q=\frac{3}{5} \times 80=48 \mu C\)
\(2 \mu F\) 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 \(q=\frac{3}{5} \times 80=48 \mu C\)
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