Question

Consider the circuit shown in thye figure above. Take Let g = 6 V, L = 6 mH, and R = 4 Ohms. s E E 000 1. R (a) Calculate the inductive time constant (b) Calculate the current in the circuit 250 ms after the switch is closed. (c) Assumed the switched is closed for a long time. What is the current in the circuit? (d) How long does it take the current to reach 80% of its maximum value?

Answer 1

Here we have given that,

E = 6 V

L = 6 mH

R = 4 ohm

A)

Inductive time constant T = L/R = 6 mH/4 = (0.0015) Sec

B) t = 250 ms

Current in the inductor as function of time will be,

I(t) = Io(1-e^-t/T)

Here Io = E/R = 6/4 = 1.5 A

So that,

I(t = 250 ms) = 1.5(1-e^-250×10^-3/1.5×10^-3)

I(t) = 1.5 A

Means here the circuit will have reached the steady state.

C) at time t = infinity,

Inductor becomes short circuited so that,

I = E/R = 6/4 =1.5 A

D) to reach 80% of maximum value of current here,

0.80Io = Io(1-e^-t/T)

On taking both side log here will get,

-t/T = ln(0.20)

t = (2.41415686865 ×10^-3 Sec) = 2.41 mSec

Which is the required time .