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Self Inductance MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Self inductance is considered one the most difficult concept.

  • 30 Questions around this concept.

Solve by difficulty

Consider the LR circuit shown in the figure. If the switch S is closed at $t=0$ then the amount of charge that passes through the battery between $t=0$ and $t=\frac{L}{R}$ is :

 

An ideal coil of 10 H is connected in series with a resistance of $5 \Omega$ and a battery of 5 V .2 second after the connection is made, the current flowing in ampere in the circuit is

Let C be the capacitance of a capacitor discharging through a resistor R . Suppose t_{1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_{2} is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_{1} / t_{2} will be

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The current (I) in the inductance is varying with time according to the plot shown in figure.

Which one of the following is the correct variation of voltage with time in the coil?

In the circuits (a) and (b) switches S1 and S2 are closed at t=0 and are kept closed for a long time. The variation of currents in the two circuits for t\geq 0 are roughly shown by (figures are schematic and not drawn to scale) :

The inductance (in H)  between A and D is

Concepts Covered - 1

Self inductance

Inductance-

It is the property of electrical circuits that oppose any change in the current in the circuits.

Inductance is analogous to inertia in mechanics.

Self Inductance-

Whenever the electric current passing through a coil or circuit changes, the magnetic flux linked with it will also change. And to oppose this flux change according to Faraday’s laws of electromagnetic induction, an emf is induced in the coil or the circuit. This
the phenomenon is called ‘self-induction’.

or Self-inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing.

And the emf induced is called back emf, current so produced in the coil is called induced current.

And the direction of induced current for case A and case B  is shown below.

Coefficient of self-induction (L)-

If $\phi$ is the flux linkages associated with 1 turn of the coil. And if N is the number of turns in the coil.
Then total flux linkage associated with the coil is $N \phi$
And this total flux linkage is directly proportional to the current in the coil. I.e $N \phi \alpha i$
or we can write $\phi_{\text {total }}=\phi_T=N \phi=L i$
where $L=$ coefficient of self-induction.

So the coefficient of self-induction is given as

$$
L=\frac{N \phi}{I}
$$

. If $i=1 \mathrm{amp}, N=1$ then, $L=\phi$
i.e The coefficient of self-induction of a coil is equal to the flux linked with the coil when the current in it is 1 amp.

Faraday Second Law of Induction emf-
Using $\phi_{\text {total }}=N \phi=L i \quad$ and $\quad \varepsilon=\frac{-d \phi_T}{d t}$
we get $\quad \varepsilon==-N \frac{d \phi}{d t}=-L \frac{d i}{d t}$

$$
\text { If } \frac{d i}{d t}=1 \frac{a m p}{s e c} \text { and } N=1 \text { then }|\varepsilon|=L
$$
 

      i.e  The coefficient of self-induction is equal to the emf induced in the coil when the rate of change of current in the coil is unity.

Units and dimensional formula of ‘L’-

S.I. Unit - Henry (H)

And $1 H=\frac{1 V \cdot s e c}{A m p}$
And its dimensional formula is $M L^2 T^{-2} A^{-2}$
Dependence of self-inductance (L)-
It depends upon the number of turns $(\mathrm{N})$, Area $(\mathrm{A})$ and permeability of medium $(\mu)$.

‘L’ does not depend upon current flowing or change in current flowing.

 

 

 

 

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Self inductance

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