JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
In the given circuit, the charge on \(4\, \mu F\) capacitor will be.....\(\mu C\)
- A \(13.4\)
- B \(24\)
- C \(9.6\)
- D \(5.4\)
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 \(\Rightarrow \quad \frac{q_{1}}{q_{2}}=\frac{2.4}{3}=\frac{4}{5}\) \(\therefore \quad 9 X=54\, \mu C\) \(\therefore \quad X=6\, \mu C\)…
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