Page 374 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 29] ALTERNATING-CURRENT CIRCUITS 359

RESONANCE
The impedance in a series RLC ac circuit is a minimum when X L = X C ; under these circumstances Z = R and
I = V/R. The resonance frequencyf 0 of a circuit is that frequency at which X L = X C :
1 1
2π f 0 L = f 0 = √
2π f 0 C 2π LC
When the potential difference applied to a circuit has the frequency f 0 , the current in the circuit is a maximum.
This condition is known as resonance. At resonance the phase angle is zero since X L = X C .

SOLVED PROBLEM 29.11
In the antenna circuit of a radio receiver that is tuned to a particular station, R = 5 , L = 5 mH, and
C = 5pF. (a) Find the frequency of the station. (b) If the potential difference applied to the circuit is
5 × 10 −4 V, find the current that flows.
1 1
(a) f 0 = √ = = 1006 kHz
2π LC 2π (5 × 10 −3 H)(5 × 10 −12 F)
(b) At resonance, X L = X C and Z = R. Hence,
V 5 × 10 −4 V −4
I = = = 10 A = 0.1mA
R 5

SOLVED PROBLEM 29.12
In the antenna circuit of Prob. 29.11, the inductance is fixed but the capacitance can be varied. What
should the capacitance be in order to receive an 800-kHz radio signal?

1 1
C = =
2
3
2
(2π f ) (L) [(2π)(800 × 10 Hz)] (5 × 10 −3 H)
= 7.9 × 10 −12 F = 7.9pF
SOLVED PROBLEM 29.13
In a series ac circuit R = 20 , X L = 10 , and X C = 25 when the frequency is 400 Hz. (a) Find the
impedance of the circuit. (b) Find the phase angle. (c) Is the resonance frequency of the circuit greater
than or less than 400 Hz? (d) Find the resonance frequency.

(a) Z = R + (X L − X C ) = (20 ) + (10 − 25 ) = 25
2
2
2
2
10 − 25
X L − X C
(b) tan φ = = =−0.75 φ =−37 ◦
R 20
A negative phase angle signifies that the voltage lags behind the current.
(c) At resonance X L = X C . At 400 Hz, X L < X C , so the frequency must be changed in such a way as to increase
X L and decrease X C . Since X L = 2π f L and X C = 1/(2π f C), it is clear that increasing the frequency will
have this effect, Hence the resonance frequency must be greater than 400 Hz.
(d) Since X L = 10 and X C = 25 when f = 400 Hz,
10
X L −3
L = = = 4 × 10 H
2π f (2π)(400 Hz)
1 1 −5
C = = = 1.6 × 10 F
2π f X C (2π)(400 Hz)(25 )
1 1
Hence f 0 = √ = = 629 Hz
2π LC 2π (4 × 10 −3 H)(1.6 × 10 −5 F)