AP EAMCET · PHYSICS · Capacitance
The charge on \(4 \mu \mathrm{F}\) capacitor, in the given circuit is
- A \(24 \mu \mathrm{C}\)
- B \(100 \mu \mathrm{C}\)
- C \(2.4 \mu \mathrm{C}\)
- D \(30 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(24 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
The circuit can be redrawn as, \(C_{\mathrm{cq}}=\frac{4 \times 6}{4+6}=2.4 \mu \mathrm{F}\) As we know, potential drop across parallel branch \(A B\) is \(V_{A B}=10 \mathrm{~V}\) So, charge \(Q=C_{\mathrm{eq}} V=2.4 \times 10 \times 10^{-6} \mathrm{C}\) \(Q=24 \mu \mathrm{C}\)…
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