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Question:

In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and τ=RC is capacitive time constant). Which of the following statement is correct?

Work done by the battery is half of the energy dissipated in the resistor.

At t=τ, q=CV/2

At t=2τ, q=CV(1−e⁻²)

At t=τ/2, q=CV(1−e⁻¹/²)

Solution:

Charge on the capacitor at any time t is given by:
q = CV(1 - e⁻ᵗ/τ)
where:
q = charge on the capacitor
C = capacitance
V = voltage of the battery
t = time
τ = time constant (RC)

Let's analyze the given options:

Option 1: Work done by the battery is half of the energy dissipated in the resistor.
The energy stored in the capacitor is (1/2)CV². The energy dissipated in the resistor is also (1/2)CV². Therefore, the work done by the battery is equal to the energy dissipated in the resistor, not half of it. This statement is incorrect.

Option 2: At t=τ, q=CV/2
Substituting t = τ in the equation for q:
q = CV(1 - e⁻¹)
q ≈ CV(1 - 0.368) ≈ 0.632CV
This statement is incorrect.

Option 3: At t=2τ, q=CV(1−e⁻²)
Substituting t = 2τ in the equation for q:
q = CV(1 - e⁻²)
This statement is correct.

Option 4: At t=τ/2, q=CV(1−e⁻¹/²)
Substituting t = τ/2 in the equation for q:
q = CV(1 - e⁻¹/²)
This statement is correct (although not the best answer because option 3 is explicitly given)

Therefore, the correct statement is that at t=2τ, q=CV(1−e⁻²).