Question

Current in a circuit falls from 5.0A to 0.0A in 0.1s. If an average emf of 200V induced, calculate the self-inductance of the circuit.

Answer 1

Faraday’s law of emf can be used to solve this problem. Subtract the final current from the initial current and substitute in the formula. Rearranging the equation and substituting values in it will give the self-inductance of the coil.

Formula used: dI= ${ I }_{ f }$-${ I }_{ i }$; $e=-L\dfrac { dI }{ dt }$

Complete step-by-step solution:

Given: Initial current (${ I }_{ i }$)= 5.0A

Final current (${ I }_{ f }$)= 0.0A

Time taken for the charge (t)= 0.1s

Average emf (e)= 200V

Change in current is given by,

dI= ${ I }_{ f }$-${ I }_{ i }$ ………….(1)

$\therefore$ dI= -5A

Relation between self-inductance and average emf from Faraday’s law is given by,

$e=-L\dfrac { dI }{ dt }$

Where, L is the self-inductance of the coil

Rearranging equation. (2) we get,

$L=\dfrac { e }{- \dfrac { dI }{ dt } }$

Substituting the values in above equation we get,

$L=\dfrac { 200 }{ \dfrac { 5 }{ 0.1 } }$

$\therefore L=4H$

Hence, the self-inductance of the coil is 4H.

Note: The self-inductance emf is also called back emf. It is called so because it opposes any change in current in the circuit. Hence, work is needed to be done against back emf for establishing current.

When the emf is induced in the same circuit in which the current is changing, then that emf is called Self-induction. It is denoted by L. But when the emf is induced in another coil situated in the same magnetic field, then that emf is called Mutual Induction. It is denoted by M.