CHAPTER 23 ALTERNATING
CURRENT CIRCUITS
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13. REASONING The rms current can be calculated from Equation 23.3, Irms = Vrms / XL ,
provided that the inductive reactance is obtained first. Then the peak value of the current I0
supplied by the generator can be calculated from the rms current Irms by using Equation
20.12, I 0 = 2 I rms .
SOLUTION At the frequency of f = 620 Hz , we find, using Equations 23.4 and 23.3, that
X L = 2π f L = 2π (620 Hz)(8.2 ×10 –3 H) = 32 Ω
I rms =
Vrms
XL
=
10.0 V
= 0.31 A
32 Ω
Therefore, from Equation 20.12, we find that the peak value I0 of the current supplied by the
generator must be
I 0 = 2 I rms = 2 ( 0.31 A ) = 0.44 A
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25. REASONING We can use the equations for a series RCL circuit to solve this problem,
provided that we set X L = 0 since there is no inductance in the circuit. Thus, according to
Equations 23.6 and 23.7, the current in the circuit is I rms = Vrms / R2 + X C2 . When the
frequency f is very large, the capacitive reactance is zero, or XC = 0 , in which case the
current becomes I rms (large f ) = Vrms / R . When the current Irms in the circuit is one-half
the value of I rms (large f ) that exists when the frequency is very large, we have
I rms
I rms (large f )
=
1
2
We can use these expressions to write the ratio above in terms of the resistance and the
capacitive reactance. Once the capacitive reactance is known, the frequency can be
determined.
SOLUTION The ratio of the currents is
I rms
I rms (large f )
=
Vrms / R 2 + X C2
Vrms / R
Taking the reciprocal of this result gives
=
R
R 2 + XC2
=
1
2
or
R2
1
=
4
R 2 + XC2
2
ALTERNATING CURRENT CIRCUITS
R 2 + XC2
= 4
R2
1+
or
X C2
R2
= 4
Therefore,
XC
R
=
3
According to Equation 23.2, XC = 1/ ( 2π f C ) , so it follows that
XC
R
=
1/ ( 2π f C )
=
R
3
Thus,
f =
1
1
=
2π RC 3
2π (85 Ω ) 4.0 ×10 –6 F
(
)
3
= 270 Hz
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37. REASONING Since we know the values of the resonant frequency of the circuit, the
capacitance, and the generator voltage, we can find the value of the inductance from
Equation 23.10, the expression for the resonant frequency. The resistance can be found
from energy considerations at resonance; the power factor is given by cosφ , where the
phase angle φ is given by Equation 23.8, tanφ = (XL – XC )/ R .
SOLUTION
a. Solving Equation 23.10 for the inductance L, we find that
L =
1
1
=
= 2.94 × 10 –3 H
2
2
3
2
–6
4π f0 C
4π (1.30 × 10 Hz) (5.10 ×10 F)
2
b. At resonance, f = f0 , and the current is a maximum. This occurs when X L = X C , so
2
that Z = R . Thus, the average power P provided by the generator is P = Vrms
/ R , and
solving for R we find
2
V rms
(11.0 V) 2
R =
=
= 4.84 Ω
P
25.0 W
c. When the generator frequency is 2.31 kHz, the individual reactances are
XC =
1
1
=
=13.5 Ω
3
2π f C 2π (2.31×10 Hz)(5.10 ×10 –6 F)
X L = 2π f L = 2π (2.31 ×10 3 Hz)(2.94 ×10 –3 H) = 42.7 Ω
Chapter 23
3
The phase angle φ is, from Equation 23.8,
 X – XC 
 42.7 Ω –13.5 Ω 
φ = tan –1 L
= tan–1 

 = 80.6°
R
4.84 Ω
The power factor is then given by
cosφ = cos 80.6° = 0.163
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45. REASONING Since the capacitor and the inductor are connected in parallel, the voltage
across each of these elements is the same or VL = VC . Using Equations 23.3 and 23.1,
respectively, this becomes Irms X L = Irms X C . Since the currents in the inductor and
capacitor are equal, this relation simplifies to X L = X C . Therefore, we can find the value of
the inductance by equating the expressions (Equations 23.4 and 23.2) for the inductive
reactance and the capacitive reactance, and solving for L.
SOLUTION Since X L = X C , we have
2π f L =
1
2π f C
Therefore, the value of the inductance is
L=
1
1
=
= 0.176 H = 176 mH
4π 2 f 2 C 4π 2 (60.0 Hz) 2 ( 40.0 ×10 –6 F)
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