__Growth of Current :__

A series combination of an inductor L and a resistor R are connected across a cell of e.m.f. E through a switch S as shown.

When switch is closed current starts increasing in the Inductor . This causes an induction of e.m.f. in the Inductor. The induced e.m.f. opposes the growth of current in the circuit.

Let at any time t current in the circuit be I_{ .}

From loop rule we obtain,

$ \displaystyle \xi = L\frac{dI}{dt} + IR $

$ \displaystyle -L\frac{dI}{dt} = IR -\xi $

$ \displaystyle \frac{dI}{IR – \xi} = -\frac{1}{L} dt $

Integrating ,

$ \displaystyle I = \frac{\xi}{R} (1 – e^{-R t/L}) $

Here, I represents the instantaneous current in the circuit.

__Decay of Current :__

In this case source of emf. is disconnected from the circuit

$ \displaystyle -L\frac{dI}{dt} – IR = 0 $

$ \displaystyle \int_{I_0}^{I} \frac{dI}{I} = -\frac{R}{L}\int_{0}^{t} dt $

$ \displaystyle I = I_0 e^{-R t/L} $

(L/R) is called time constant as its dimension is same as that of time.

Example : A current of I = 10 A is passed through the part of a circuit shown in the figure. What will be the potential difference between A and B when I is decreased at constant rate of 10^{2} amp/s, at the beginning?

Solution : Applying the law of potential between the points A and B we obtain,

V_{B} − V_{A} = −IR + E −L di/dt

=> V_{B} − V_{A} = −10 × 2 + 12 − 5 × 10^{-3} × 10^{2}

=> V_{B} − V_{A} = −20 + 12 − 0.5

=> V_{B} − V_{A} = −8.5 volt.