AP EAMCET · PHYSICS · Capacitance
In the circuit shown in figure, if the point \(R\) is earthed and point \(P\) is given a potential of \(+1800 \mathrm{~V}\), then charges on \(C_2\) and \(C_3\) are respectively
- A \(2.4 \times 10^{-3} \mathrm{C} ; 1.2 \times 10^{-3} \mathrm{C}\)
- B \(1.6 \times 10^{-3} \mathrm{C} ; 0.8 \times 10^{-3} \mathrm{C}\)
- C \(3.2 \times 10^{-3} \mathrm{C}: 1.6 \times 10^{-3} \mathrm{C}\)
- D \(4.8 \times 10^{-3} \mathrm{C} ; 2.4 \times 10^{-3} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(2.4 \times 10^{-3} \mathrm{C} ; 1.2 \times 10^{-3} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(\begin{gathered}C_{\mathrm{eq}} \text { of system }=\left(C_2 \text { parallel } C_3\right) \text { series } C_1 \\ =1 /\left(\frac{1}{3}+\frac{1}{(4+2)}\right)=2 \mu \mathrm{F}\end{gathered}\) So, charge taken from source…
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