An LR circuit having a time constant of 50 ms is connected with an ideal battery of emf ε. find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value and (c) the magnetic field energy stored in the circuit reaches half its maximum value.

#### Solution

Given:-

Time constant of the LR circuit = 50 ms

Emf of the battery = ε

The time constant of the LR circuit is given by

\[\tau = \frac{L}{R} = 50 ms = 0 . 05 s\]

Let the current reach half of its maximum value in time t.

Now,

\[\frac{i_0}{2} = i_0 (1 - e^{- t/0 . 05} )\]

\[ \Rightarrow \frac{1}{2} = 1 - e^{- t/0 . 05} \]

\[ \Rightarrow e^{- t/0 . 03} = \frac{1}{2}\]

On taking natural logarithm (ln) on both sides, we get

\[\ln e^{- t/0 . 05} = \ln\left( \frac{1}{2} \right)\]

\[ \Rightarrow - \frac{t}{0 . 05} = \ln(1) - \ln(2)\]

\[ \Rightarrow - \frac{t}{0 . 05} = 0 - 0.6931\]

\[ \Rightarrow t = 0 . 05 \times 0 . 6931\]

\[ = 0 . 03465 s\]

\[ = 35 ms\]

(b) Let t be the time at which the power dissipated is half its maximum value.

Maximum power = `E^2/R`

\[\therefore \frac{E^2}{2R} = \frac{E^2}{R}(1 - e^{tR/L} )^2 \]

\[ \Rightarrow 1 - e^{- R/L} = \frac{1}{\sqrt{2}} = 0 . 707\]

\[ \Rightarrow e^{- tR/L} = 0 . 293\]

\[ \Rightarrow t = 50 \times 1 . 2275 ms\]

\[ = 61 . 2 ms\]

(c) Current in the coil at the steady state, `i = epsilon/R`

Magnetic field energy stored at the steady state,

\[U = \frac{1}{2}L i^2\text{ or }U\]

\[= \frac{\epsilon^2}{2 R^2}L\]

Half of the value of the steady-state energy = \[\frac{1}{4}L\frac{\epsilon^2}{R^2}\]

Now,

\[\frac{1}{4}L\frac{\epsilon^2}{R^2} = \frac{1}{2}L\frac{\epsilon^2}{R^2}(1 - e^{- tR/L} )^2 \]

\[ \Rightarrow e^{- tR/L} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2}\]

\[ \Rightarrow t = \tau\left[ \ln\left( \frac{1}{2 - \sqrt{2}} \right) + \ln 2 \right]\]

\[= 0 . 05\left[ \ln\left( \frac{1}{2 - \sqrt{2}} \right) + \ln 2 \right]\]

\[= 0 . 061 s\]

\[= 61 ms\]