In a fluorescent lamp choke (a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be ________.

Solution

Physics sir · 4 steps

The induced emf tells you how hard the choke fights the current change: divide it by the rate of change.

1. Induced emf. A changing current in a coil induces an emf that opposes the change. Its size is the self-inductance times the rate of change of current.

   |e| = L (Δ I)/(Δ t)

2. Rate of change. The current falls by Δ I = 0.25 A in Δ t = 0.025 ms = 2.5 × 10⁻⁵ s. So (Δ I)/(Δ t) = 0.25/(2.5 × 10⁻⁵) = 1 × 10⁴ A s⁻¹.
3. Solve for L. L = |e|/(Δ I/Δ t) = 100/(1 × 10⁴) = 0.01 H.
4. So. L = 0.01 H = 10 mH, so the answer is 10.

The common slip: 0.01. Leaving the answer in henry gives 0.01. The question asks for mH, so multiply by 1000 to get 10.

The idea to keep. Self-inductance is just emf divided by dI/dt. Put the time in seconds first, find L in henry, and only then convert to the unit the question asks for.

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