Question

A 24 V battery is connected in series with a resistor and an inductor, with R = 11.0 Ω and L = 10.0 H, respectively. Find the energy stored in the inductor for the following situations:

(a) when the current reaches its maximum value
J

(b) one time constant after the switch is closed
J

Answer 1

The Inductor behaves as open for a sudden change in current, hence it acts as "open" when t=0+ or just after closing the switch.

The Inductor behaves as "Short" when t=∞ .

a) When Inductor becomes short in RL circuit, The current will be maximum and will be equal to V/R, i.e., at t=∞;

The energy stored in an inductor when current reaches maximum is E= (1/2) L i²

i= V/R = 24/11 = 2.181 A

E = (1/2) x 10 x(2.181)² =23.783 jouls

b) time constant , τ = L/R = 10/11 = 0.909

Hence the velue of current after one time constant will be

i = (I_{final-Iinitial)(}e^-(t/τ))

i = (2.181-0) (1-e^-1) = 1.378 A

Energy stored in inductor after one time constant = E= (1/2) L i²

E = (1/2) x10x(1.378)² = 9.4944 jouls