Question

An inductance l and a resistance r are connected in series with a battery of emf the maximum rate at which the energy is stored in the magnetic field is

Answer 1

The maximum rate at which the energy is stored in the magnetic field is E =($i^2$ R +P $\frac{(\frac{di}{dt})}{i}$).

Explanation:

• using the law of dealing with voltage the equation is as follows:

                              E - iR-L  = 0

• To convert it into a power equation we have to multiply the equation with i

                             Ei - $i^2$ R- Li $\frac{di}{dt}$ = 0.

• This shows that the total power is equal to the sum of our stored and power dissipated.  

• The magnetic field has stored energy E equal to $\frac{1}{2} Li^2$

• On differentiating the magnetic field energy we will get the maximum rate of energy stored, $\frac{d}{dt} (\frac{1}{2}Li^2)$ = Li = P

Thus,

                               Ei =P  $(\frac{di}{dt})$ + $i^2$ R

                              ⇒  E =( $i^2$ R +P $\frac{(\frac{di}{dt})}{i}$ )

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Maximum rate at which energy is stored in magnetic field

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Answer 2

Solution :

• The energy stored in the magnetic field at time t is

$\bf U = \frac{1}{2} {Li}^{2} = \frac{1}{2} {Li_0}^{2} {(1 - {e}^{ - t/ \tau} )}^{2}$

• The rate at which the energy is stored as

$\bf p = \frac{du}{dt} = {Li_0}^{2} (1 - {e}^{ - t \tau} )( - {e}^{ - t \tau}) \bigg( - \frac{1}{ \tau} \bigg) \\ \\ \bf \implies \frac{ {Li_0}^{2} }{ \tau} ( {e}^{ - t/ \tau} - {e}^{ - 2t \tau} ).....(i)$

• This rate will be maximum whem dP/dt = 0

$\bf \frac{ {Li_0}^{2} }{ \tau} \bigg( - \frac{1}{ \tau} {e}^{ - t/ \tau} + \frac{2}{ \tau} {e}^{ - 2t/ \tau} \bigg) = 0 \\ \\ \bf \implies {e}^{ - t/ \tau} = \frac{1}{2}$

Putting in ( i ),

$\bf P_{max} = \frac{ {Li_0}^{2} }{ \tau} \bigg( \frac{1}{2} - \frac{1}{4}\bigg) \\ \\ \bf \implies \frac{L{\epsilon}^{2}}{4{R}^{2}(L/R)} \\ \\ \large \bf \orange{ \implies \frac{{\epsilon}^{2}}{4R}}$