Q: In the figure shown is a R-L circuit connected with a cell of emf ξ through a key k . If key k is closed find the current drawn by the battery

(a) just after the key k is closed

(b) long after the key k is closed

Solution : (a) In the process of charging, the current in the inductor is given as $\displaystyle I = I_0 (1-e^{-t/\tau}) $

where I_{0} = steady state current through the inductor & τ = L/R_{eq }; R_{eq} = equivalent resistance of R-L circuit

Therefore at t = 0, i = 0,

No current passes through the inductor just after closing the switch therefore the inductor should be eliminated initially as it carries no current initially. Total initial current drawn from the battery.

$\displaystyle I_0 = \frac{\xi}{R_0}$ ; Where R_{0} = the equivalent resistance between the point P & Q.

referring the following figure we conclude that

R_{0} = 2R

(b) After a long time (t → ∞) the current passing through the inductor is given as

$\displaystyle I = I_0 (1-e^{-t/\tau}) = I_0 $

If is equal to the steady state current that means after a very long time the inductor behaves as zero resistance. Since the inductor has zero resistance to D.C. (steady state current) the equivalent resistance between R & S is equal to zero,

The total resistance between P & Q is given as

R_{0} = 5R/3

The steady state current $\displaystyle = \frac{\xi}{R_0} = \frac{\xi}{5R/3}$