Page 356 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 28] ELECTROMAGNETIC INDUCTION 341

INDUCTORS IN COMBINATION
When two or more inductors are sufficiently far apart for them not to interact electromagnetically, their equivalent
inductances when they are connected in series and in parallel are as follows:
L = L 1 + L 2 + L 3 +· · · inductors in series
1 1 1 1
= + + +· · · inductors in parallel
L L 1 L 2 L 3
Connecting coils in parallel reduces the total inductance to less than that of any of the individual coils. This is
rarely done because coils are relatively large and expensive compared with other electronic components; a coil
of the required smaller inductance would normally be used in the first place.
Because the magnetic field of a current-carrying coil extends beyond the inductor itself, the total inductance
of two or more connected coils will be changed if they are close to one another. Depending on how the coils are
arranged, the total inductance may be larger or smaller than if the coils were farther apart. This effect is called
mutual inductance and is not considered in the above formula.

SOLVED PROBLEM 28.12

Find the equivalent inductances of a 5- and an 8-mH inductor when they are connected (a) in series and
(b) in parallel.

(a) L = L 1 + L 2 + L 3 = 5mH + 8mH = 13 mH
1 1 1 1 1
(b) = + = + L = 3.08 mH
L L 1 L 2 5mH 8mH

ENERGY OF A CURRENT-CARRYING INDUCTOR
Because a self-induced emf opposes any change in an inductor, work has to be done against this emf to establish
a current in the inductor. This work is stored as magnetic potential energy. If L is the inductance of an inductor,
its potential energy when it carries the current I is

1
W = LI 2
2
This energy powers the self-induced emf that opposes any decrease in the current through the inductor.

SOLVED PROBLEM 28.13
(a) How much magnetic potential energy is stored in a 20-mH coil when it carries a current of 0.2 A?
(b) What should the current in the coil be in order that it contains1Jof energy?
1
2
1
2
(a) W = LI = ( )(20 × 10 −3 H)(0.2A) = 4 × 10 −4 J
2 2

2W (2)(1J)
(b) I = = = 10 A
L 20 × 10 −3 H
TIME CONSTANT
Because the self-induced emf in a circuit such as that of Fig. 28-5 is always such as to oppose any changes in the
current in the circuit, the current does not rise instantly to its final value of I = V/R when the switch is closed.
When the switch in Fig. 28-5 is closed, the current starts to build up; as a result, a self-induced emf −L( I/ t)
occurs that opposes the battery voltage V . The net voltage in the circuit is therefore V − L( I/ t), which must