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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2019
  • Hard
  • Numerical

In the given circuit, the charge on \(4\, \mu F\) capacitor will be.....\(\mu C\)

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Verified Solution

Answer & Solution

Correct Answer

(B) \(24\)

Step-by-step Solution

Detailed explanation

So total charge flow \(=q=5.4\,\mu F \times 10\, V\)

\(=54\, \mu \mathrm{F}\)

The charge will be distributed in the ratio of capacitance \(\Rightarrow \quad \frac{q_{1}}{q_{2}}=\frac{2.4}{3}=\frac{4}{5}\)

\(\therefore \quad 9 X=54\, \mu C\)

\(\therefore \quad X=6\, \mu C\)

Charge on \(4 \,\mu F\) capacitor will be \(=4 \mathrm{X}=4 \times 6\, \mu \mathrm{C}\)

\(=24\, \mu C\)

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