8.0 Growth and Decay of current in an LR circuit:

In the figure, a resistor $R$, an inductor $L$ and a source of emf $\varepsilon $ connected in series through a switch $S$. Switch $S$ is open so there is no current in circuit initially. At $t=0$ switch is closed, now current start to flow. As the current is increase in circuit, self induced emf is produced.

By Kirchhoff’s loop law,

$$\varepsilon - L\frac{{di}}{{dt}} = Ri$$

$$L\frac{{di}}{{dt}} = \varepsilon - Ri$$

$$\frac{{di}}{{\varepsilon - Ri}} = \frac{{dt}}{L}$$

At $t = 0,i = 0$ and at time t the current is $i$. Thus,

$$\int\limits_0^i {\frac{{di}}{{\varepsilon - Ri}}} = \int\limits_0^t {\frac{{dt}}{L}} $$

or, $$ - \frac{1}{R}\ln \frac{{\varepsilon - Ri}}{\varepsilon } = \frac{t}{L}$$

$$\frac{{\varepsilon - Ri}}{\varepsilon } = {e^{ - tR/L}}$$

$$\varepsilon - Ri = \varepsilon {e^{ - tR/L}}$$

$$i = \frac{\varepsilon }{R}(1 - {e^{ - tR/L}}).$$

The constant $\frac{L}{R}$ is called the time constant of the $LR$ circuit. $\frac{L}{R} = \tau $ and $\frac{\varepsilon }{R} = {i_0}$.

$$i = {i_0}(1 - {e^{ - t/\tau }}).$$

Consider the arrangement shown in figure. The switch $S$ can be switched to $a$ from $b$. The circuit is complete and $i=0$ is maintained through the circuit. Now at $t=0$ switch is moved from $b$ to $a$ position (assuming that it took zero time to move switch from $b$ position to $a$). Circuit is complete through $Aa$ and battery is disconnect to the circuit. As battery is disconnected through circuit, the current is start to decrease. Hence emf is induced in circuit. So,

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

$$\frac{{di}}{i} = - \frac{R}{L}dt$$

At $t = 0,i = {i_0}$. If the current at time t be $i$,

$$\int\limits_{{i_0}}^i {\frac{{di}}{i}} = \int\limits_0^t { - \frac{R}{L}dt} $$

$$\ln \frac{i}{{{i_0}}} = - \frac{R}{L}t$$

$$i = {i_0}{e^{ - tR/L}}$$

$$i = {i_0}{e^{ - t/\tau }}$$

where $\tau $ is the time constant of the circuit. Here we see that current does not fall to zero. It gradually decrease as time passes.