R – L Circuit : Growth & Decay of Current

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 102 amp/s, at the beginning?

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

VB − VA = −IR + E −L di/dt

=> VB − VA = −10 × 2 + 12 − 5 × 10-3 × 102

=> VB − VA = −20 + 12 − 0.5

=> VB − VA = −8.5 volt.