When an ac source v is connected to a pure inductor, Kirchhoff’s loop rule gives \( v-L\frac{di}{dt}=0 \). The second term, \( -L\frac{di}{dt} \), is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the self-induced emf
By Kirchhoff’s loop rule, the emfs around the loop add up to zero. A circuit with only a source and a pure inductor has two emfs. They are the source voltage \( v \) and the emf that the changing current induces in the inductor.
Option (1) the mutually induced emf — Incorrect. A mutually induced emf appears in one coil because of a changing current in a different, nearby coil. Here there is only one coil, and the changing current is its own.
Option (2) the self-induced emf — Correct. As the current \( i \) in the inductor changes, the magnetic flux linked with the same coil changes, and an emf \( -L\,di/dt \) is induced in it. This is the self-induced emf, and its negative sign follows from Lenz’s law. It opposes the change in current.
Option (3) the power dissipated in the circuit — Incorrect. \( L\,di/dt \) has the unit henry × ampere per second, which is the volt. So the term is an emf. A power is measured in watts.
NCERT (2026-27):
AC Voltage Applied to an Inductor – “… Let the voltage across the source be v = vm sinωt. Using the Kirchhoff’s loop rule, \( \sum\varepsilon(t)=0 \), and since there is no resistor in the circuit,
\( \displaystyle v-L\frac{di}{dt}=0 \qquad (7.10) \)
where the second term is the self-induced Faraday emf in the inductor; and L is the self-inductance of the inductor. The negative sign follows from Lenz’s law (Chapter 6).”