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In an L-C-R circuit, both switches are initially open. The capacitor (capacitance C) is connected to a battery of EMF V through a resistor R by closing switch S1 while S2 stays open, so the capacitor charges. Let q be the charge on the capacitor and tau = RC the time constant. Which statement is correct?

1. Work done by the battery is half of the energy dissipated in the resistor
2. At t = tau, q = C*V/2
3. At t = 2*tau, q = C*V*(1 - e⁻²)
4. At t = tau/2, q = C*V*(1 - e⁻¹)

Correct answer: At t = 2*tau, q = C*V*(1 - e⁻²)

Solution

For RC charging, q(t) = CV(1 - e^(-t/tau)). At t = 2tau, q = CV(1 - e⁻²), which matches that option. At t = tau, q = CV(1 - e⁻¹) approx 0.63CV, not CV/2, so that option is wrong. At t = tau/2, q = CV(1 - e⁻⁰.5), not CV(1 - e⁻¹). Energy-wise, work by battery = CV² (split equally: half stored, half dissipated), so dissipated energy equals stored energy, not 'work = half of dissipated'. So the correct statement is q = CV(1 - e⁻²) at t = 2tau.

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