Page 372 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 29] ALTERNATING-CURRENT CIRCUITS 357

Fig. 29-8

(a) The reactance of the capacitor at 60 Hz is
1 1
X C = = = 531
2π f C (2π)(60 Hz)(5 × 10 −6 F)
Since X L = 0, the impedance of the circuit is (Fig. 29-8)

2 2 2 2
Z = R + (X L − X C ) = R + (−X C )

2
2
2
2
= R + X = (300 ) + (531 ) = 610
C
Hence the current is
V 120 V
I = = = 0.197 A
Z 610
0 − 531
X L − X C
(b) tan φ = = =−1.77
R 300
φ =−61 ◦
The negative phase angle signifies that the current in the circuit leads the voltage.
SOLVED PROBLEM 29.9
A 5-mH, 20- inductor is connected to a 28-V, 400-Hz power source. Find (a) the current in the inductor
and (b) the phase angle.

(a) The reactance of the inductor is
X L = 2π f L = (2π)(400 Hz)(5 × 10 −3 H) = 12.6
and its impedance is (Fig. 29-9)

2
2
2
2
Z = R + X = (20 ) + (12.6 ) = 23.6
L

Fig. 29-9