Q136 / 150 · MHT CET 2024 (09 May Shift 1) · Physics

In the given circuit diagram, in the steady state the current through the battery and the charge on the capacitor respectively are Figure

  • A
    2 A and $3 \mu \mathrm{C}$
  • B
    $\frac{6}{11} \mathrm{~A}$ and $\frac{12}{7} \mu \mathrm{C}$
  • Correct answer
    C
    11 A and $3 \mu \mathrm{C}$
  • D
    zero ampere and $3 \mu \mathrm{C}$

Solution

In steady state, current through capacitor is zero, hence current through $4 \Omega$ resistor becomes zero. $\begin{array}{ll} \therefore & \frac{1}{R_{e q}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6} \\ \therefore & I=\frac{V}{R_{e q}}=\frac{6 \times 11}{6}=11 \mathrm{~A} \end{array}$ $\therefore \quad$ Charge on capacitor is, $\mathrm{Q}_{\mathrm{C}}=\mathrm{CV}=0.5 \times 6=3 \mu \mathrm{C}$

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

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