Ch. 24 Solution
PHY 205
H.S.
Chapter 24: Alternating-Current Circuits
Answers Conceptual Questions
4.
As the frequency is increased, the inductive reactance increases as well. Therefore, at
frequencies greater than the resonance frequency of an LC circuit, the inductive reactance is
greater than the capacitive reactance. As a result, the inductor dominates, and the voltage
leads the current. This means that the phase angle, φ, is positive.
10.
Yes. All that is required for their resonance frequencies to be the same is for the product of L
and C to be the same. (See equation 24-18.)
Answers Conceptual Exercises
6.
At low frequency, the capacitor is essentially the same as a break in the circuit, whereas the
inductor is essentially an ideal wire. It follows, then, that more current will be supplied by
the generator if the inductor and the capacitor are connected in parallel.
10.
At high frequency, we can replace the inductor with an open circuit. At low frequency, we
can replace the capacitor with an open circuit. In either case, the effective resistance of the
circuit is R; therefore, the current is the same at both high and low frequency.
Solutions to Problems
4. Picture the Problem: Power is dissipated at a rate proportional to the square of the current in a
resistive AC circuit.
Strategy: Solve equation 21-5 for the average power dissipated, using the rms current. To calculate the
peak power dissipated, replace the rms current with the peak current, using equation 24-3.
Solution: 1. (a) Use equation 21-5 to Pav = I rms 2 R = ( 0.85 A ) (150 Ω ) = 110 W = 0.11 kW
find Pav :
2
2. (b) Calculate the peak power
dissipated:
Pmax = I max 2 R =
(
2 I rms
)
2
R = 2 ( I rms 2 R ) = 2 (110 W ) = 0.22 kW
Insight: When the current oscillates as a sine wave, the average power dissipation is one-half the peak
power dissipation.
14. Picture the Problem: An oscillating voltage drives an alternating
current through a capacitor as in the circuit depicted at the right.
Strategy: Insert equation 24-9 for the capacitive reactance into
equation 24-8 and solve for the rms current.
Solution: 1. (a)
Calculate the rms
current at f =
1.00 kHz:
I rms =
Vrms
V
= rms = ω CVrms = 2π (1.00 kHz )( 0.395 µ F )(10.0 V ) = 24.8 mA
X C 1 ωC
2. (b) Since the current is directly proportional to the frequency of the voltage, doubling the frequency
doubles the current.
1
Ch. 24 Solution
3. (c) Calculate
the rms current
at f = 2.00
kHz:
PHY 205
H.S.
I rms = 2π ( 2.00 kHz )( 0.395 µ F )(10.0 V ) = 49.6 mA
Insight: The rms current is proportional to the oscillation frequency.
24. Picture the Problem: An oscillating voltage drives an alternating
current through both a capacitor and a resistor that are connected in
series.
Strategy: Calculate the phase shift from equation 24-12, where the
impedance is given by equation 24-11 and the capacitive reactance
by equation 24-9. Using the phase shift, sketch the phasor diagram
for the voltage across the resistor, capacitor, and power source.
2
Calculate the power dissipated in the resistor using Pav = I rms
R.
Write the rms current as the rms voltage divided by the impedance.
Solution: 1. (a) Calculate
XC :
2. Calculate the impedance:
X C = 1 ω C = ⎡⎣ 2π ( 60.0 Hz ) ( 32.2 ×10−6 F ) ⎤⎦ = 82.4 Ω
−1
Z = R 2 + X C2 =
3. Calculate the phase angle: φ = cos −1
(105 Ω )
2
+ ( 82.4 Ω ) = 133 Ω
2
R
105 Ω
= cos −1
= 37.9°
Z
133 Ω
4. Sketch the phasor diagram with the current leading the voltage
by φ = 37.9°. The resistive vector and the capacitive vector
should have lengths proportional to their 105-Ω and 82.4-Ω
magnitudes, respectively:
2
5. (b) Calculate
the power
dissipated in the
resistor:
⎛V ⎞
Pav = I rms 2 R = ⎜ rms ⎟ R
⎝ Z ⎠
2
⎛ 120 V ⎞
=⎜
⎟ (105 Ω ) = 85.5 W
⎝ 133 Ω ⎠
Insight: In the phasor diagram the voltage across the resistor is in phase with the current. The voltage
in the capacitor lags 90° behind the current, and the voltage produced by the power source lags the
current by 37.9°.
40. Picture the Problem: An oscillating voltage drives an alternating current through both an inductor and
a resistor that are connected in series in an air conditioner.
Strategy: Use equation 24-15 to calculate the impedance of the circuit. Divide the voltage by the
impedance to calculate the rms current (Ohm’s Law). Multiply the current squared times the resistance
to calculate the average power consumed by the air conditioner.
Solution: 1. (a) Calculate the impedance: Z = R 2 + X L 2 =
( 7.0 Ω )
2
+ (15 Ω ) = 16.6 Ω = 17 Ω
2
Vrms 240 V
=
= 14.46 A = 14 A
Z
16.6 Ω
2. (b) Divide the voltage by the
impedance:
I rms =
3. (c) Calculate the average power
consumed:
Pav = I rms 2 R = (14.46 A ) 7.0 Ω = 1.5 kW
2
2
Ch. 24 Solution
PHY 205
H.S.
Insight: The reactance in the circuit limits the current, but does not consume power. The power factor
for this air conditioner is cos φ = 0.412.
48. Picture the Problem: The figure shows a 2.50-Ω resistor, 0.300mH inductor, and 0.100-µF capacitor connected in series with a
5.00-V, 30.0-kHz power supply.
Strategy: When the voltmeter is connected between the points A
and B, it measures the rms voltage across the inductor. When
connected between points B and C, it measures the rms voltage
across the capacitor. When connected across points A and C, it
measures the rms voltage of the capacitor and inductor. When
connected
across points A and D, it measures the rms voltage of the power source. Calculate the rms voltage
across each element by multiplying the impedance of that element by the rms current. To calculate the
rms current, divide the rms voltage by the impedance of the entire circuit, which is given by equation
24-16.
1
1
=
= 53.05 Ω
ω C 2π ( 30.0 kHz )( 0.100 µ F )
Solution: 1. Calculate the
capacitive reactance:
XC =
2. Calculate the inductive
reactance:
X L = ω L = 2π ( 30.0 kHz )( 0.300 mH ) = 56.55 Ω
3. Calculate the impedance:
Z = R2 + ( X L − X C ) =
= 4.3 Ω
4. Calculate the rms current:
I rms =
5. (a) Multiply
the current by
the inductive
reactance:
Vrms,L = I rms X L = (1.4 A )( 56.55 Ω ) = 79 V
6. (b) Multiply
the current by
the capacitive
reactance:
Vrms,C = I rms X C = (1.4 A )( 53.05 Ω ) = 74 V
7. (c) Calculate
the impedance
between points
A and C:
Z LC =
8. Multiply the current by the
impedance:
Vrms,LC = I rms Z LC = (1.4 A )( 3.5 Ω ) = 4.9 V
9. (d) Multiply the current by the total
impedance:
Vrms = I rms Z = (1.4 A )( 4.3 Ω ) = 6.0 V
2
( 2.50 Ω ) + ( 56.55 − 53.05 Ω )
2
2
Vrms 6.00 V
=
= 1.4 A
Z
4.3 Ω
( X L − XC )
2
= 56.55 − 53.05 Ω = 3.5 Ω
Insight: Note that the voltages across the inductor and across the capacitor are much greater than the
voltage across the power source. Since the inductor and capacitor are 180 degrees out of phase with
each other, they tend to cancel out each other’s instantaneous voltage.
58. Picture the Problem: An RLC circuit consists of a 105-Ω resistor, 518-mH inductor, and a 0.20-µF
capacitor connected in series with an oscillating power source.
3
Ch. 24 Solution
PHY 205
H.S.
Strategy: Use equation 24-18 to calculate the resonance frequency. Then insert the resistance,
inductance, and new capacitance into equation 24-16 to determine the impedance. Divide the resistance
by the impedance to calculate the power factor.
Solution: 1. (a)
Calculate the
resonance
frequency:
2. (b) Calculate
the impedance
with the larger
capacitor:
3. (c) Divide
the resistance
by the
impedance:
f =
1
2π LC
=
1
2π
( 0.518 H ) ( 0.200 ×10− 6 F )
1 ⎞
⎛
Z = R2 + ⎜ ω L −
⎟
ω
C⎠
⎝
= 494 Hz
2
⎡
⎤
1
= (105 Ω ) + ⎢ 2π ( 494.47 Hz )( 0.518 H ) −
⎥
2π ( 494.47 Hz ) 0.220 µ F ⎥⎦
⎢⎣
= 180 Ω
2
2
cos φ =
R 105 Ω
=
= 0.58
Z 180 Ω
Insight: In this circuit, increasing the capacitance by only 10% decreased the power factor by over
40%. There are only two significant figures in the answers because
X L − X C = 1.61 − 1.46 kΩ = 0.15 kΩ in step 2.
4