Q149 / 200 · Manipal (MET) 2011 · Physics

A capacitor of capacitance $5 \mu \mathrm{~F}$ is connected as shown in the figure. The internal resistance of the cell is $0.5 \Omega$. The amount of charge on the capacitor plate is Figure

  • Correct answer
    A
    $10 \mu \mathrm{C}$
  • B
    $5 \mu \mathrm{C}$
  • C
    $6 \mu \mathrm{C}$
  • D
    $10.2 \mu \mathrm{C}$

Solution

We know that capacitor offers infinite resistance for DC source. Current taken from the cell $\begin{aligned} & i=\frac{E}{R+r} \\ \therefore \quad i & =\frac{2.5}{1+1+0.5}=1 \mathrm{~A}\end{aligned}$ Potential drop across capacitor $\begin{aligned} V & =E-i r \\ & =2.5-1 \times 0.5=2 \mathrm{~V}\end{aligned}$ The current in $2 \Omega$ resistor is zero. Charge on capacitor $\begin{aligned} Q & =C V \\ & =5 \times 2=10 \mu \mathrm{C}\end{aligned}$

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

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