Page 353 - Schaum's Outline of Theory and Problems of Applied Physics

338 ELECTROMAGNETIC INDUCTION [CHAP. 28

primary and secondary windings, so the ratio of turns in the winding determines the ratio of voltages across them:
V 1 N 1
=
V 2 N 2
Primary voltage primary turns
=
Secondary voltage secondary turns
Since the power I 1 V 1 going into a transformer must equal the power I 2 V 2 going out, where I 1 and I 2 are the
primary and secondary currents, respectively, the ratio of currents is inversely proportional to the ratio of turns:
I 1 N 2
=
I 2 N 1
Primary voltage secondary turns
=
Secondary voltage primary turns

SOLVED PROBLEM 28.4

Alternating current is in wide use chiefly because its voltage can be so easily changed by a transformer.
Since P = IV , we see that for given power P the higher the voltage, the lower the current, and vice versa.
In transmitting electric energy through long distances, a small current is desirable in order to minimize
2
energy loss to heat, which is equal to I R where R is the resistance of the transmission line. However,
both the generation and the final use of electric energy are best accomplished at moderate potential
differences. Hence electricity is typically generated at 10,000 V or so, stepped up by transformers at
the power station to 500,000 V or even more for transmission, and near the point of consumption other
transformers reduce the potential difference to 240 or 120 V. To verify the advantage of high-voltage
transmission, find the rate of energy loss to heat when a 5- cable is used to transmit 1 kW of electricity
at 100 V and at 100,000 V.
Since P = IV , the currents in the cable are, respectively,
P 1000 W P 1000 W
I A = = = 10 A I B = = = 0.01 A
V A 100 V V B 100,000 V
The rates of heat production per kilowatt are, respectively,
2
2
I R = (10 A) (5 ) = 500 W
A
2
2
I R = (0.01 A) (5 ) = 0.0005 W = 5 × 10 −4 W
B
6
Transmission at 100 V therefore means 10 —1 million—times more energy lost as heat than does transmission at
100,000 V.
SOLVED PROBLEM 28.5
A transformer has 100 turns in its primary winding and 500 turns in its secondary winding. If the primary
voltage and current are, respectively, 120 V and 3 A, what are the secondary voltage and current?

N 2 500 turns
V 2 = V 1 = (120 V) = 600 V
N 1 100 turns

N 1 100 turns
I 2 = I 1 = (3A) = 0.6A
N 2 500 turns
SOLVED PROBLEM 28.6
A transformer rated at a maximum power of 10 kW is used to connect a 5000-V transmission line to a
240 V circuit. (a) What is the ratio of turns in the windings of the transformer? (b) What is the maximum
current in the 240-V circuit?