Page 377 - Schaum's Outline of Theory and Problems of Applied Physics

362 ALTERNATING-CURRENT CIRCUITS [CHAP. 29

1 1
(a) f 0 = √ = = 41 Hz
2π LC 2π (0.3H)(50 × 10 −6 F)
(b) At the resonance frequency, X L = X C and Z = R. Hence
V 120 V
I = = = 1.5A
R 80
R R
(c) cos φ = = = 1 = 100%
Z R
(d) True power = IV cos φ = (1.5A)(120 V)(1) = 180 W

(e) Apparent power = IV = (1.5A)(120 V) = 180 VA

SOLVED PROBLEM 29.18

(a) Find the potential difference across the resistor, the inductor, and the capacitor in the circuit of Prob.
29.15 when it is connected to a 120-V ac source whose frequency is equal to the circuit’s resonance
frequency of 41 Hz. (b) Are these values in accord with the applied potential difference of 120 V?

(a) At the resonance frequency of f 0 = 41 Hz, the inductive and capacitive reactances are, respectively,

X L = 2π f 0 L = (2π)(41 Hz)(0.3H) = 77
1 1
X C = = = 77
2π f 0 C (2π)(41 Hz)(50 × 10 −6 F)
The various potential differences are therefore
V R = IR = (1.5A)(80 ) = 120 V

V L = IX L = (1.5A)(77 ) = 116 V
V C = IX C = (1.5A)(77 ) = 116 V

(b) The total potential difference across the circuit is

2 2 2 2
V = V + (V L − V C ) = (120 V) + (0) = 120 V
R
which is equal to the applied potential difference.
SOLVED PROBLEM 29.19

A 5-hp electric motor is 80 percent efficient and has an inductive power factor of 75 percent. (a) What
minimum rating in kilovoltamperes must its power source have? (b) A capacitor is connected in series
with the motor to raise the power factor to 100 percent. What minimum rating in kilovoltamperes must
the power source now have?
(a) The power required by the motor is

(5hp)(0.746 kW/hp)
P = = 4.66 kW
0.8
Since P = IV cos φ, the power source must have the minimum rating

P 4.66 kW
IV = = = 6.22 kVA
cos φ 0.75
(b) When cos φ = 1, IV = P = 4.66 kVA.