A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V. Determine approximately :
(a) The time elapsed before the current acquires half of its steady - state value.
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given ln 2=0.693, e^(-3 / 2)=0.25)

Solution

Physics sir · 4 steps

Find the time constant L/R first. The half-time and the 15 ms current both come straight from it.

1. Growth of current. In an LR circuit switched on at t = 0, the current rises as I = I₀(1 - e^(-t/τ)). Here τ = L/R = 1/100 = 10 ms and I₀ = 6/100 = 0.06 A.
2. Half the steady value. Put I = I₀/2. Then e^(-t/τ) = 1/2, so t = τ ln 2 = 10 × 0.693 = 6.93 ms ≈ 7 ms.
3. Current at 15 ms. Here t/τ = 15/10 = 3/2, and e^(-3/2) = 0.25 is given. So I = 0.06 × (1 - 0.25) = 0.045 A.

   U = 1/2LI² = 1/2 × 1 × (0.045)² = 1.01 × 10⁻³ J

4. So. t ≈ 7 ms and U ≈ 1 mJ, which is option (c).

The idea to keep. In an LR circuit, everything is measured in units of τ = L/R. Half the final current always takes τ ln 2, and the stored energy uses the current at that instant, not the final current.

Answer from the NTA final answer key. Solution by PrepWiser.