The charge on the capacitor of capacitance $C$ shown in the figure below will be

  • A
    $CE$
  • B
    $\frac{CE R}{R_1+r}$
  • C
    $\frac{C E R_2}{R_2+r}$
  • D
    $\frac{C E R_1}{R_2+r}$

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The stored charge in the capacitor in steady state of the following circuit is . . . . . . $\mu C$.

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$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

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