An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :

Solution

Physics sir · 4 steps

The stored energy gives you L, the heating gives you R, and the time constant is just L over R.

1. Inductance from energy. Magnetic energy stored is U = 1/2LI². So 64 = 1/2 × L × 8² = 32L, which gives L = 2 H.
2. Resistance from power. The coil dissipates energy only through its resistance, so P = I²R. Then 640 = 8² × R = 64R, which gives R = 10 Ω.

   τ = L/R = (2 H)/(10 Ω) = 0.2 s

3. Quick check. Dividing the two formulas gives τ = L/R = 2U/P = (2 × 64)/640 = 0.2 s. The current cancels, so the answer does not depend on the 8 A.
4. So. The time constant of the coil across an ideal battery is 0.2 s, option (d).

The idea to keep. A real coil is an L and an R in series. Energy 1/2LI² fixes L, heating I²R fixes R, and τ = L/R = 2U/P. Never forget the half in the energy formula.

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