Q121 / 160 · WBJEE 2011 · Physics

The charge on the capacitor of capacitance \(\mathrm{C}\) shown in the figure below will be Figure

  • A
    \(\mathrm{CE}\)
  • B
    \(\frac{\mathrm{CE} \mathrm{R}}{\mathrm{R}_1+r}\)
  • Correct answer
    C
    \(\frac{C E R_2}{R_2+r}\)
  • D
    \(\frac{C E R_1}{R_2+r}\)

Solution

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 \(\mathrm{Q}=\mathrm{CV}=\frac{\mathrm{CER}_2}{\mathrm{R}_2+\mathrm{r}}\)

Hence, the correct answer is option: 'C'.

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