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Energy Stored In An Inductor MCQ - Practice Questions with Answers

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

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  • Energy stored in an inductor is considered one the most difficult concept.

  • 15 Questions around this concept.

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An inductor coil of inductance L is divided into two equal parts and both parts are connected in parallel. The net inductance is :

The equivalent inductance between points P and Q in figure is :

The dimension of $\sqrt{\frac{L}{\mathrm{C}}}$ is same as that of:

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Shown in the figure is a R-L-C circuit. In steady state the ratio of energy stored in the inductor & capacitor is equal to

An inductor coil stores 32 J of magnetic energy and dissipates energy as heat at the rate of 320 W when a current of 4A is passed through it. What is the time constant of the circuit when this coil is joined across an ideal battery?

A solenoid of inductance 100 mH and resistance 20\Omega is connected to a cell of emf 10 V. What is the energy stored in the inductor when the time t = 5 ln 2 milli sec?

A constant current flows in an R-L circuit. Then

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A network of inductances, each of value 1H, is shown in figure. The equivalent inductance of the circuit between points A and B is

The magnetic potential energy stored in a certain inductor is 25mJ. When the current in the inductor is 60mA. This inductor is of inductance 

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Concepts Covered - 1

Energy stored in an inductor

Energy stored in an inductor (U)-

In building a steady current in the circuit, the source emf has to do work against of self-inductance of the coil and whatever energy
consumed for this work stored in the magnetic field of coil this energy called as magnetic potential energy (U) of the coil.

When an electric current i is flowing in an inductor, there is energy stored in the magnetic field. Considering a pure inductor L, the instantaneous power which must be supplied to initiate the current in the inductor is

P=i v=L i \frac{d i}{d t}

The work done by the voltage source during a time interval $dt$ is

d W=P d t= i L \frac{d i}{d t} d t=L i d i

total work $W$ done in establishing the final current $I$ in the inductor

\text { W }=\int_{0}^{t} P d t=\int_{0}^{I} L i d i =\frac{1}{2} L I^{2}

So Energy stored in the magnetic field of the inductor is given as

U= \frac{1}{2}LI^{2}

 

The energy density (u)/Energy per unit volume-

using U= \frac{1}{2}LI^{2}

for the solenoid field, we can write 

  U=\frac{1}{2}(L i) i=\frac{N \phi i}{2} 

u=\frac{U}{V}= \frac{B^{2}}{2\mu _{0}}

 

 

 

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Energy stored in an inductor

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