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JEE Main 2021 (1 Sep, Shift 2) · Capacitors in DC Circuits & RC

Capacitors in DC Circuits & RC question from JEE Main 2021 (1 Sep, Shift 2)

Q28JEE Main 2021 · 1 SEP · SHIFT 2Medium

Capacitors in DC Circuits & RC

A capacitor is connected to a 20 V battery through a resistance of 10Ω\Omega. It is found that the potential difference across the capacitor rises to 2 V in 1 μ\mus. The capacitance of the capacitor is __________ μ\muF. Given: ln⁡(109)=0.105\ln\left(\frac{10}{9}\right)=0.105

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Answer: (b)

  1. Charging: V=V0(1−e−t/RC)V = V_0(1-e^{-t/RC}), so 2=20(1−e−t/RC)⇒e−t/RC=0.92 = 20(1-e^{-t/RC}) \Rightarrow e^{-t/RC} = 0.9.
  2. tRC=ln⁡109=0.105\dfrac{t}{RC} = \ln\dfrac{10}{9} = 0.105
  3. C=t0.105R=10−60.105×10=0.95×10−6 FC = \dfrac{t}{0.105R} = \dfrac{10^{-6}}{0.105\times10} = 0.95\times10^{-6}\text{ F}
  4. So the answer is (b) 0.95 μ\muF.
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