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JEE Main 2017 · Physics · Electromagnetic Induction and Alternating Current · hard

A capacitor of 4 muF, initially uncharged, is charged through a resistor of 500 kΩ by a battery of emf 100 V. The time taken from the instant of…

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A capacitor of 4 μF4\ \mu\mathrm{F}, initially uncharged, is charged through a resistor of 500 kΩ500\ \mathrm{k\Omega} by a battery of emf 100 V100\ \mathrm{V}. The time taken from the instant of connection for the energy stored in the capacitor to reach half of its final value is nearly (use ln⁡2=0.693\ln 2 = 0.693, 2=1.414\sqrt2 = 1.414)
  1. A. 2.46 s2.46\ \mathrm{s}
  2. B. 1.39 s1.39\ \mathrm{s}
  3. C. 2.00 s2.00\ \mathrm{s}
  4. D. 4.62 s4.62\ \mathrm{s}
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Answer: (A) 2.46 s2.46\ \mathrm{s}

The time constant is τ=RC=5×105×4×10−6=2 s\tau = RC = 5\times10^5\times4\times10^{-6} = 2\ \mathrm{s}. The charge grows as q=q0(1−e−t/τ)q = q_0(1 - e^{-t/\tau}) and the energy as q2/2Cq^2/2C. Half the final energy 12⋅q022C\tfrac12\cdot\dfrac{q_0^2}{2C} needs q=q0/2q = q_0/\sqrt2, so 1−e−t/τ=0.70711 - e^{-t/\tau} = 0.7071, e−t/τ=0.2929e^{-t/\tau} = 0.2929 and t=τln⁡(3.414)=2×1.228≈2.46 st = \tau\ln(3.414) = 2\times1.228 \approx 2.46\ \mathrm{s}. Half the charge (not half the energy) would be reached at τln⁡2=1.39 s\tau\ln2 = 1.39\ \mathrm{s}.
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Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake