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.....\(\mu C\)

Pick an option to check your answer.
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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