Download CHAPTER 33 ALTERNATING

CHAPTER 33
ALTERNATING-CURRENT CIRCUITS
Problem
20. A 2.0-  F capacitor has a capacitive reactance of 1.0 k. (a) What is the frequency of the applied
voltage? (b) What inductance would give the same value for inductive reactance at this frequency? (c)
How would the two reactances compare if the frequency were doubled?
Solution
(a) From Equation 33-5, f  [2 XC C]1  [2 (1 k)(2  F)]1  79.6 Hz. (b) X L  XC implies
L  X C /  (1 k)  (2  10 3 s)  2 H (see Equation 33-7 and the calculation in part (a)). (c)
Doubling  doubles X L and halves XC , so X L would be four times X C at f  159 Hz.
Problem
22. A voltage V  Vp sin  t is applied across a capacitor. What is the minimum frequency for which the
current will be zero at time t  20  s?
Solution
The instantaneous current in a capacitor subjected to a voltage V (t )  Vp sin  t is I (t )  Vp C cos  t (see
previous solution). I(t)  0 for
t  20  s is 2 ft  / 2, or
t   /2, 3 /2, etc. The smallest frequency satisfying this condition at
f  1/4t  (4  20 s) 1  12.5 kHz.
Problem
36. In an LC circuit, what fraction of a cycle passes before the energy in the capacitor falls to one-fourth
of its peak value?
Solution
The energy stored in the capacitor of an LC circuit is U E  U E, p cos 2  t (see “Analyzing the LC Circuit”
in Section 33-3), where U E , p is the peak value. Then
 t  (2 /T )t  cos 1 ( 12 )   /3 (or 60),
so
U E /U E , p  1/4  cos2 t, implies
t/T 
1
6
cycle.
Problem
42. A series RLC circuit has R  75 , L  20 mH, and a resonant frequency of 4.0 kHz. (a) What is the
capacitance?
(b) What is the impedance of the circuit at resonance? (c) What is the impedance at 3.0 kHz?
Solution
(a) From the expression for resonance in an RLC circuit,
C  1/02 L  (2  4 kHz) 2 (20 mH) 1  79.2 nF. (b) At resonance, X L  XC  0, so
Z  R  75 . (c) At 3 kHz,
X L  X C   L  (1/C )  (2  3 kHz  20 mH) 
(2  3 kHz  79.2 nF)1  293 , and Z 
R2  ( XL  XC )2  (75 )2  (293 )2  303 .
Problem
54. A series RLC circuit has resistance 100  and impedance 300 . (a) What is the power factor? (b) If
the rms current is 200 mA, what is the power dissipation?
Solution
(a) The power factor is cos   R/Z  100 /300   0.333 (see solution to the next problem). (b) A
combination of the result in part (a) with Equations 33-14 and 17 shows that
2
P  I rmsVrms cos   I rms ( I rms Z )( R/Z )  I rms
R  (200 mA)2 (100 )  4 W. (Note that the average
AC power is given by the same expression as the DC power if the rms current is used.)
Problem
58. A rural power line carries 2.3 A rms at 4000 V. A stepdown transformer reduces this to 235 V rms to
supply a house. Find (a) the turns ratio of the transformer and (b) the current in the 235-V line to the
house.
Solution
(a) The turns ratio given in Equation 33-18 is N sec /N pri  Vsec /Vpri  235/4000  1/17. (b) If there are
no transformer losses, Equation 33-19 gives I sec  (Vpri /Vsec ) I pri  (17)(2.3 A)  39.1 A (rms).