StreakPeaked· Practice

ExamsJEE MainPhysics

A 12 V battery connected to a coil of resistance 6 Ω through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The induced emf across the coil is 20 V. The inductance of the coil is

  1. 10 mH
  2. 8 mH
  3. 5 mH
  4. 12 mH

Correct answer: 10 mH

Solution

The induced emf in an inductor is given by the formula emf = -L (di)/(dt). Given the induced emf of 20 V and the change in current over the time interval of 1 ms, we can rearrange the formula to find the inductance L. Substituting the values leads to an inductance of 10 mH.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →