Page 359 - Schaum's Outline of Theory and Problems of Applied Physics

344 ELECTROMAGNETIC INDUCTION [CHAP. 28

SOLVED PROBLEM 28.16
A 0.2-H inductor with a resistance of 3 is connected to a 6-V battery whose internal resistance is 1 .
(a) Find the final current in the circuit. (b) Find the current in the circuit 0.01, 0.05, and 0.1 s after the
connection is made.

(a) The total resistance in the circuit is the sum of the 3- resistance of the inductor and the 1- internal resistance
of the battery, so

R = 3 + 1 = 4
The final current in the circuit is therefore
V 6V
I 0 = = = 1.5A
R 4
(b) The time constant of the circuit is
L 0.2H
T = = = 0.05 s
R 4
At time t after the connection is made, the current is given by
I = I 0 (1 − e −t/T )

When t = 0.01 s, t/T = (0.01 s)/(0.05 s) = 0.2. With the help of a calculator or a table of exponentials, we
find that

e −t/T = e −0.2 = 0.82
and so I = I 0 (1 − e −t/T ) = (1.5A)(1 − 0.82) = (1.5A)(0.18) = 0.27 A

When t = 0.05 s, t = T and

I = 0.63I 0 = (0.63)(1.5A) = 0.95 A
When t = 0.1s, t/T = (0.1s)/(0.05 s) = 2, and
e −t/T = e −2 = 0.14

and so I = (I 0 )(I − e −t/T ) = (1.5A)(1 − 0.14) = (1.5A)(0.86) = 1.29 A

SOLVED PROBLEM 28.17
Find the current in the above inductor 0.01 and 0.1 s after it has been short-circuited, after having been
connected to the battery for a long time.

The resistance of the circuit is now just the 3- resistance of the inductor itself. Hence the time constant is
L 0.2H
T = = = 0.067 s
R 3
When t = 0.01 s, t/T = (0.01 s)/(0.067 s) = 0.15, and

e −t/T = e −0.15 = 0.86
Hence the current in the inductor is

I = I 0 e −t/T = (1.5A)(0.86) = 1.29 A
When t = 0.1s, t/T = (0.1s)(0.067 s) = 1.5, and

e −t/T = e −1.5 = 0.22