Question

2 points
4. The energy stored in coil
carrying current is u. If
current is halved, then energy
stored in the coil will be
O (a) U/2
O (b) U/4
0 (c) 20
O (d) 4U​

Answer 1

Answer:.

If current is halved (I/2), then energy stored in the coil will be one fourth of energy stored in coil carrying I.

Therefore, option (b) is correct.

Explanation:

Suppose a current is applied coil of inductance L such that current through the coil grows from zero value to a maximum value I. Let at any instant of time, current through the inductor to be 'i' . A emf induced in the coil which opposes the flow of current  is :-

$E =-L\frac{di}{dt}$

Where di/dt is the rate change of current and L is inductance.

The work done is :-  $dW = -Eidi$

$dW=-(-L\frac{di}{dt} )idt$

$dW=iLdi$

Integrating both sides:-

$\int\limits^W_0 {dW}=\int\limits^I_0 {iL} \, di$

$W=L\int\limits^I_0 {idi}$

$W=L[\frac{i^2}{2} ]^I_o$

$W=\frac{1}{2} LI^2$

The energy (U) stored in the coil when current , I is equal to:-

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

When the current halved, the energy (U') stored in the coil:-

$U'=\frac{1}{2}L(\frac{I}{2} )^2$

$U'=\frac{1}{8}LI^2$                                                                  

$U'=\frac{1}{4}(\frac{1}{2}LI^2 )$

$U'=\frac{U}{4}$

Therefore, the energy stored (U') in the coil will be one fourth of U.

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