Page 373 - Schaum's Outline of Theory and Problems of Applied Physics

358 ALTERNATING-CURRENT CIRCUITS [CHAP. 29

Hence the current is
V 28 V
I = = = 1.18 A
Z 23.6
12.6 − 0
X L − X C
(b) tan φ = = = 0.63
R 20
φ = 32 ◦
The positive phase angle signifies that the voltage in the circuit leads the current.

SOLVED PROBLEM 29.10

A 10-µF capacitor, a 0.10-H inductor, and a 60- resistor are connected in series across a 120-V, 60-Hz
power source. Find (a) the current in the circuit and (b) the phase angle.

(a) The reactances are
X L = 2π f L = (2π)(60 Hz)(0.10 H) = 38
1 1
X C = = = 265
2π f C (2π)(60 Hz)(10 × 10 −6 F)
The impedance is therefore (Fig. 29-10)

2
2
2
2
Z = R + (X L − X C ) = (60 ) + (38 − 265 ) = 235
Hence, the current in the circuit is
V 120 V
I = = = 0.51 A
Z 235
X L − X C 38 − 265 227
(b) tan φ = = =− =−3.78
R 60 60
φ =−75 ◦
The negative phase angle signifies that the current in the circuit leads the voltage.

Fig. 29-10