AP EAMCET · PHYSICS · Capacitance
A capacitor of capacitance \(C_1=10 \mu \mathrm{F}\) to charged using \(9 \mathrm{~V}\) battery. It is then removed from the battery and connected to another capacitor \(C_2=20 \mu \mathrm{F}\) as shown in the figure.
The charge on \(C_2\) after equilibrium has reached is

- A \(6.0 \times 10^{-5} \mathrm{C}\)
- B \(60 \times 10^{-6} \mathrm{C}\)
- C \(3.0 \times 10^{-5} \mathrm{C}\)
- D \(3.0 \times 10^{-6} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(6.0 \times 10^{-5} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} C_1 & =10 \mu \mathrm{F} \\ & =10^{-5} \mathrm{~F} \\ V_1 & =9 \mathrm{~V} \end{aligned} \] Charge on capacitor \(C_1\),…
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