Page 381 - Schaum's Outline of Theory and Problems of Applied Physics

366 ALTERNATING-CURRENT CIRCUITS [CHAP. 29

(b) The phasor diagram of Fig. 29-14(b) shows how the various currents are to be added. We have

2 2 2 2
I = I + (I C − I L ) = (1.0A) + (0.5A − 0.8A)
A
√

2
2
2
2
2
= (1.0A) + (−0.3A) = 1.00 A + 0.09 A = 1.09 A = 1.04 A
V 10 V
(c) Z = = = 9.6
I 1.04 A
I R 1.0A
(d) cos = = 0.962
I 1.04 A
φ =−16 ◦
◦
The current lags behind the voltage by 16 , as in Fig. 29-14(b):
P = IV cos φ = (10 V)(1.04 A)(0.962) = 10 W
We can also find P in this way:
2
2
P = I R = (1.0A) (10 ) = 10 W
R

IMPEDANCE MATCHING
In Chapter 25 we saw that power transfer between two dc circuits is a maximum when their resistances are equal,
an effect called impedance matching. The same effect occurs in an ac circuit with the impedance of the circuits
instead of their resistances as the quantities that must be the same for maximum power transfer. An additional
consideration is that an impedance mismatch between ac circuits may result in a distorted signal by altering the
properties of the circuits, for instance, by changing their resonance frequencies.
A transformer enables impedances between ac circuits to be matched. An example is an audio system in
which the impedance of the amplifier circuit might be several thousand ohms whereas the impedance of the voice
coil of the loudspeaker might be only a few ohms. Connecting the voice coil directly to the amplifier would be
extremely inefficient; using a transformer is the answer. The ratio of turns N 1 /N 2 between the windings of the
transformer for maximum power transfer from a circuit of impedance Z 1 to a circuit of impedance Z 2 is

N 1 Z 1
=
N 2 Z 2

SOLVED PROBLEM 29.23
√
Derive the formula N 1 /N 2 = Z 1 /Z 2 for maximum power transfer.
From Chapter 28

V 1 N 1 I 2 N 1
= and =
V 2 N 2 I 1 N 2
Because Z 1 = V 1 /I 1 and Z 2 = V 2 /I 2 , the impedance ratio is
2

Z 1 V 1 I 2 N 1
= =
Z 2 V 2 I 1 N 2
The ratio of turns needed to match the impedances is therefore

N 1 Z 1
=
N 2 Z 2