Page 354 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 28] ELECTROMAGNETIC INDUCTION 339

N 1 V 1 5000 V
(a) = = = 20.8
N 2 V 2 240 V
(b) Since P = IV , here
P 10,000 W
I 2 = = = 41.7A
V 2 240 V

SOLVED PROBLEM 28.7
A transformer connected to a 120-V alternating-current (ac) power line has 200 turns in its primary
winding and 50 turns in its secondary winding. The secondary is connected to a 100- light bulb. How
much current is drawn from the 120-V power line?
The voltage across the secondary is

N 2 50 turns
V 2 = V 1 = (120 V) = 30 V
N 1 200 turns
and so the current in the secondary circuit is
V 2 30 V
I 2 = = = 0.3A
R 100
Hence, the current in the primary circuit is

N 2 50 turns
I 1 = I 2 = (0.3A) = 0.075 A
N 1 200 turns

SELF-INDUCTANCE
When the current in a circuit changes, the magnetic field enclosed by the circuit also changes, and the resulting
change in flux leads to a self-induced emf of
I
Self-induced emf = V =−L
t
Here I/ t is the rate of change of the current, and L is a property of the circuit called its self-inductance,
or, more commonly, its inductance. The minus sign indicates that the direction of V is such that it opposes the
change in current I that created it.
The unit of inductance is the henry (H). A circuit or circuit element that has an inductance of 1 H will have
a self-induced emf of 1 V when the current through it changes at the rate of 1 A/s. Because the henry is a rather
large unit, the millihenry and microhenry are often used, where

−3
1 millihenry = 1mH = 10 H
1 microhenry = 1 µH = 10 −6 H

A circuit element with inductance is called an inductor. A solenoid is an example of an inductor. The
inductance of a solenoid is
2
µN A
L =
l
where µ is the permeability of the core material, N is the number of turns, A is the cross-sectional area, and l is
the length of the solenoid (Fig. 28-4).