Question

An inductor $L=2\text{H}$ and a resistance $R=8\Omega$ are connected in series with a battery of emf $E=8\text{V}.$ The maximum rate at which the energy is stored in the magnetic field is $\left(\text{in W}\right)$.(integer only)

Answer 1

Given:
$L=2\text{H};R=8\Omega ;E=8\text{V}$

The energy stored in the magnetic field at time $t$ is,

$U=\frac{1}{2}L{i}^2=\frac{1}{2}L{i}_0^2{(1-e^{\frac{-t}{\tau }})}^2.......\left(1\right)$

$[\because i=i_0(1-e^{\frac{-t}{\tau }})]$

The rate at which the energy is stored is, $P=\frac{dU}{dt}$

Therefore, differentiating the (1) we get

$P=\frac{dU}{dt}=2\times \frac{1}{2}L{i}_0^2(1-e^{\frac{-t}{\tau }})(-e^{\frac{-t}{\tau }})(-\frac{1}{\tau })$

$P=\frac{L{i}_0^2}{\tau }(e^{\frac{-t}{\tau }}-e^{\frac{-2t}{\tau }})......\left(2\right)$

The rate at which energy stored will be maximum when $\frac{dP}{dt}=0$

Therefore, differentiating eq. $\left(2\right)$ and equating to zero, we get

$\frac{dP}{dt}=\frac{L{i}_0^2}{\tau }(-\frac{1}{\tau }{e}^{\left(\frac{-t}{\tau }\right)}+\frac{2}{\tau }{e}^{\left(\frac{-2t}{\tau }\right)})=0$

$\Rightarrow e^{\frac{-t}{\tau }}=\frac{1}{2}$

Putting this value in (2),

$P_{max}=\frac{L{i}_0^2}{\tau }(\frac{1}{2}-\frac{1}{4})$

$P_{max}=\frac{L{E}^2}{4R^2\left(\frac{L}{R}\right)}=\frac{E^2}{4R}.......\left(3\right)$

Putting the values in (3) we get,

$P_{max}=\frac{E^2}{4R}=\frac{4^2}{8}$

$\therefore P_{max}=\frac{16}{8}=2\text{W}$

Note: $P_{max}$ is independent of the inductance of the circuit.