WBJEE · Physics · Capacitance
The charge on the capacitor of capacitance \(\mathrm{C}\) shown in the figure below will be
- A \(\mathrm{CE}\)
- B \(\frac{\mathrm{CE} \mathrm{R}}{\mathrm{R}_1+r}\)
- C \(\frac{C E R_2}{R_2+r}\)
- D \(\frac{C E R_1}{R_2+r}\)
Answer & Solution
Correct Answer
(C) \(\frac{C E R_2}{R_2+r}\)
Step-by-step Solution
Detailed explanation
Hints: \(I=\frac{E}{R_2+r}\) (Since finally no current flows through capacitor) \(\therefore\) Potential difference across \(\mathrm{R}_2, \mathrm{~V}=\mathrm{R}_2=\frac{\mathrm{ER}_2}{\mathrm{R}_2+\mathrm{r}}\) \(\therefore\) Charge on the capacitor…
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