HOMEWORK #11
Chapter 29
5
•
If the frequency in the circuit in Figure 29-28 is doubled, the capacitive
reactance of the circuit will (a) double, (b) not change, (c) halve,
(d) quadruple.
Determine the Concept The capacitive reactance of an capacitor varies with the
frequency according to X C = 1 ωC . Hence, doubling ω will halve XC. (c) is
correct.
The impedances of motors, transformers, and electromagnets include both
18 ••
resistance and inductive reactance. Suppose that phase of the current to a large industrial
plant lags the phase of the applied voltage by 25° when the plant is under full operation
and using 2.3 MW of power. The power is supplied to the plant from a substation 4.5 km
from the plant; the 60 Hz rms line voltage at the plant is 40 kV. The resistance of the
transmission line from the substation to the plant is 5.2 Ω. The cost per kilowatt-hour to
the company that owns the plant is $0.14, and the plant pays only for the actual energy
used. (a) Estimate the resistance and inductive reactance of the plant’s total load. (b)
Estimate the rms current in the power lines and the rms voltage at the substation. (c) How
much power is lost in transmission? (d) Suppose that the phase that the current lags the
phase of the applied voltage is reduced to 18º by adding a bank of capacitors in series
with the load. How much money would be saved by the electric utility during one month
of operation, assuming the plant operates at full capacity for 16 h each day? (e) What
must be the capacitance of this bank of capacitors to achieve this change in phase angle?
Picture the Problem We can find the resistance and inductive reactance of the
plant’s total load from the impedance of the load and the phase constant. The
current in the power lines can be found from the total impedance of the load the
potential difference across it and the rms voltage at the substation by applying
Kirchhoff’s loop rule to the substation-transmission wires-load circuit. The power
2
Rtrans . We can find the cost
lost in transmission can be found from Ptrans = I rms
savings by finding the difference in the power lost in transmission when the phase
angle is reduced to 18°. Finally, we can find the capacitance that is required to
reduce the phase angle to 18° by first finding the capacitive reactance using the
definition of tanδ and then applying the definition of capacitive reactance to find
C.
Rtrans = 5.2 Ω
εsubstation
∼
f = 60 Hz
ε rms= 40 kV
Z
δ = 25°
(a) Relate the resistance and
inductive reactance of the plant’s
total load to Z and δ:
R = Z cos δ
and
X L = Z sin δ
Express Z in terms of the rms current
I rms in the power lines and the rms
Z=
Express the power delivered to the
Pav = ε rms I rms cos δ
voltage ε rms at the plant:
ε rms
I rms
plant in terms of εrms, I rms , and δ and
and
solve for I rms :
I rms =
Substitute to obtain:
Z=
Substitute numerical values and
evaluate Z:
Substitute numerical values and
evaluate R and XL:
Z=
Pav
ε rms cos δ
2
ε rms
cos δ
Pav
(40 kV )2 cos 25° = 630 Ω
2.3 MW
R = (630 Ω ) cos 25° = 571Ω
= 0.57 kΩ
and
X L = (630 Ω )sin 25° = 266 Ω
= 0.27 kΩ
(b) Use equation (1) to find the
current in the power lines:
I rms =
2.3 MW
= 63.4 A
(40 kV )cos 25°
= 63 A
Apply Kirchhoff’s loop rule to the
circuit:
ε sub − I rms Rtrans − I rms Z tot = 0
Solve for εsub:
ε sub = I rms (Rtrans + Z tot )
Substitute numerical values and
ε sub = (63.4 A )(5.2 Ω + 630 Ω )
evaluate εsub:
= 40.3 kV
(1)
(c) The power lost in transmission is:
2
Ptrans = I rms
Rtrans = (63.4 A ) (5.2 Ω )
2
= 20.9 kW = 21kW
(d) Express the cost savings ∆C in
terms of the difference in energy
consumption (P25° − P18°)∆t and the
per-unit cost u of the energy:
Express the power lost in
transmission when δ = 18°:
∆C = (P25° − P18° )∆tu
Find the current in the transmission
lines when δ = 18°:
I18° =
Evaluate P18° :
P18° = (60.5 A ) (5.2 Ω ) = 19.0 kW
P18° = I182 ° Rtrans
2.3 MW
= 60.5 A
(40 kV ) cos 18°
2
Substitute numerical values and evaluate ∆C:
d   $0.14 
 h 
∆C = (20.9 kW − 19.0 kW )16   30
 = $128

 d   month   kW ⋅ h 
(e) The required capacitance is given
by:
Relate the new phase angle δ to the
inductive reactance XL, the reactance
due to the added capacitance XC, and
the resistance of the load R:
Substituting for XC yields:
C=
1
2πfX C
tan δ =
C=
XL − XC
⇒ X C = X L − R tan δ
R
1
2πf ( X L − R tan δ )
Substitute numerical values and evaluate C:
C=
1
= 33 µF
2π 60 s (266 Ω − (571Ω ) tan 18°)
(
-1
)
22 •
An inductor has a reactance of 100 Ω at 80 Hz. (a) What is its inductance? (b)
What is its reactance at 160 Hz?
Picture the Problem We can use X L = ωL to find the inductance of the inductor
at any frequency.
XL
2πf
(a) Relate the reactance of the
inductor to its inductance:
X L = ωL = 2πfL ⇒ L =
Solve for and evaluate L:
L=
(b) At 160 Hz:
X L = 2π 160 s −1 (0.199 H ) = 0.20 kΩ
100 Ω
= 0.199 H = 0.20 H
2π 80 s −1
(
)
(
)
27 ••
A circuit consists of two ideal ac generators and a 25-Ω resistor, all connected
in series. The potential difference across the terminals of one of the generators is given by
V1 = (5.0 V) cos(ωt – α), and the potential difference across the terminals of the other
generator is given by V2 = (5.0 V) cos(ωt + α), where α = π/6. (a) Use Kirchhoff’s loop
rule and a trigonometric identity to find the peak current in the circuit. (b) Use a phasor
diagram to find the peak current in the circuit. (c) Find the current in the resistor if α =
π/4 and the amplitude of V2 is increased from 5.0 V to 7.0 V.
Picture the Problem We can use the trigonometric identity
cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ )
to find the sum of the phasors V1 and V2 and then use this sum to express I as a
function of time. In (b) we’ll use a phasor diagram to obtain the same result and in
(c) we’ll use the phasor diagram appropriate to the given voltages to express the
current as a function of time.
(a) Applying Kirchhoff’s loop rule to
the circuit yields:
V1 + V2 − IR = 0
Solve for I to obtain:
I=
V1 + V2
R
Use the trigonometric identity cos θ + cos φ = 2 cos 12 (θ + φ ) cos 12 (θ − φ )
to find V1 + V2:
V1 + V2 = (5.0 V )[cos(ωt − α ) + cos(ωt + α )] = (5 V )[2 cos 12 (2ωt )cos 12 (− 2α )]
= (10 V )cos
π
6
cos ωt = (8.66 V ) cos ωt
Substitute for V1 + V2 and R to obtain:
I=
(8.66 V )cos ωt = (0.346 A ) cos ωt
25 Ω
= (0.35 A ) cos ωt
where
I peak = 0.35 A
(b) Express the magnitude of the
current in R:
I =
r
V
R
r
V2
The phasor diagram for the voltages
is shown to the right.
r
V
30°
30°
r
V1
r
Use vector addition to find V :
r
r
V = 2 V1 cos 30° = 2(5.0 V ) cos 30°
= 8.66 V
r
Substitute for V and R to obtain:
I =
8.66 V
= 0.346 A
25 Ω
and I = (0.35 A ) cos ωt
where
I peak = 0.35 A
r
V2
(c) The phasor diagram is shown to
the right. Note that the phase angle
r
r
between V1 and V2 is now 90°.
α
r
V
o
90 − α
δ
r
V1
Use the Pythagorean theorem to
r
find V :
Express I as a function of t:
r
r2 r 2
V = V1 + V2 =
(5.0 V )2 + (7.0 V )2
= 8.60 V
r
V
cos(ωt + δ )
R
where
δ = 45° − (90° − α ) = α − 45°
I=
 7 .0 V 
 − 45°
= tan −1 
 5.0 V 
= 9.462° = 0.165 rad
Substitute numerical values and
evaluate I:
I=
=
8.60 V
cos(ωt + 0.165 rad )
25 Ω
(0.34 A )cos(ωt + 0.17 rad )
36 ••
A two conductor transmission line simultaneously carries a superposition of
two voltage signals, so the potential difference between the two conductors is given by V
= V1 + V2, where V1 = (10.0 V) cos(ω1t) and
V2 = (10.0 V) cos(ω2t), where ω1 = 100 rad/s and ω2 = 10 000 rad/s. A 1.00 H inductor
and a 1.00 kΩ shunt resistor are inserted into the transmission line as shown in Figure 2931. (Assume that the output is connected to a load that draws only an insignificant
amount of current.) (a) What is the voltage (Vout) at the output of the transmission line?
(b) What is the ratio of the low-frequency amplitude to the high-frequency amplitude at
the output?
Picture the Problem We can express the two output voltage signals as the
product of the current from each source and R = 1.00 kΩ. We can find the
currents due to each source using the given voltage signals and the definition of
the impedance for each of them.
V1, out = I1 R
(a) Express the voltage signals
observed at the output side of the
and
transmission line in terms of the
V2, out = I 2 R
potential difference across the
resistor:
Evaluate I1 and I2:
I1 =
V1
=
Z1
(10.0 V )cos100t
= (9.95 mA ) cos100t
2
2
-1
(1.00 kΩ ) + [(100 s )(1.00 H )]
I2 =
V2
=
Z2
(10.0 V )cos10 4 t
= (0.995 mA ) cos10 4 t
2
2
4 −1
(1.00 kΩ ) + [(10 s )(1.00 H )]
and
Substitute for I1 and I2 to obtain:
V1, out = (1.00 kΩ )(9.95 mA ) cos100t
=
(9.95 V ) cos100t
where ω1 = 100 rad/s and
ω2 = 10 000 rad/s.
and
V2, out = (1.00 kΩ )(0.995 mA ) cos10 4 t
=
(b) Express the ratio of V1,out to
V2,out:
V1, out
V2, out
=
(0.995 V )cos104 t
9.95 V
= 10 : 1
0.995 V
64 ••
A series RLC circuit that has an inductance of 10 mH, a capacitance of 2.0 µF,
and a resistance of 5.0 Ω is driven by an ideal ac voltage source that has a peak emf of
100 V. Find (a) the resonant frequency and (b) the root-mean-square current at
resonance. When the frequency is 8000 rad/s, find (c) the capacitive and inductive
reactances, (d) the impedance, (e) the root-mean-square current, and
(f) the phase angle.
Picture the Problem We can use ω0 = 1 LC to find the resonant frequency of
the circuit, I rms = ε rms R to find the rms current at resonance, the definitions of XC
and XL to find these reactances at ω = 8000 rad/s, the definitions of Z and Irms to
find the impedance and rms current at ω = 8000 rad/s, and the definition of the
phase angle to find δ.
(a) Express the resonant frequency
ω0 in terms of L and C:
ω0 =
Substitute numerical values and
evaluate ω0:
ω0 =
1
LC
1
(10 mH )(2.0 µF)
= 7.1 × 10 3 rad/s
(b) Relate the rms current at
resonance to εrms and the impedance
of the circuit at resonance:
(c) Express and evaluate XC and
XL at ω = 8000 rad/s:
I rms =
ε rms = ε max
R
2R
100 V
=
2 (5.0 Ω )
= 14 A
XC =
1
1
=
−1
ωC 8000 s (2.0 µF)
(
)
= 62.50 Ω = 63 Ω
and
X L = ωL = 8000 s −1 (10 mH ) = 80 Ω
(
(d) Express the impedance in terms
of the reactances, substitute the
results from (c), and evaluate Z:
)
Z = R 2 + (X L − X C )
2
(5.0 Ω )2 + (80 Ω − 62.5 Ω )2
=
= 18.2 Ω = 18 Ω
(e) Relate the rms current at
ω = 8000 rad/s to εrms and the
impedance of the circuit at this
frequency:
(f) δ is given by:
Substitute numerical values and
evaluate δ:
I rms =
ε rms = ε max
Z
2Z
=
100 V
2 (18.2 Ω )
= 3 .9 A
 XL − XC 

R


δ = tan −1 
 80 Ω − 62.5 Ω 
 = 74°
5 .0 Ω


δ = tan −1 
75
••
A resistor and a capacitor are connected in parallel across an ac generator
(Figure 29-43) that has an emf given by ε = εpeak cos ωt. (a) Show that the current in the
resistor is given by IR = (εpeak/R) cos ωt. (b) Show that the current in the capacitor branch
is given by IC = (εpeak/XC) cos(ωt + 90º). (c) Show that the total current is given by I =
Ipeak cos(ωt + δ), where tan δ = R/XC and
Ipeak = εpeak/Z.
Picture the Problem Because the resistor and the capacitor are connected in
parallel, the voltage drops across them are equal. Also, the total current is the sum
of the current through the resistor and the current through the capacitor. Because
these two currents are not in phase, we use phasors to calculate their sum. The
amplitudes of the applied voltage and the currents are equal to the magnitude of the
r
r
r
r
phasors. That is ε = ε peak , I = I peak , I R = I R , peak , and I C = I C , peak .
(a) The ac source applies a voltage
given by ε = ε peak cos ωt . Thus, the
voltage drop across both the load
resistor and the capacitor is:
ε peak cos ωt = I R R
The current in the resistor is in phase
with the applied voltage:
I R = I R , peak cos ωt
Because I R , peak =
ε peak
R
:
(b) The current in the capacitor leads
the applied voltage by 90°:
Because I C , peak =
ε peak
XC
:
(c) The net current I is the sum of the
currents through the parallel branches:
IR =
ε peak
R
cos ωt
I C = I C, peak cos(ωt + 90°)
IC =
ε peak
XC
I = I R + IC
cos(ωt + 90°)
Draw the phasor diagram for the
circuit. The projections of the phasors
onto the horizontal axis are the
instantaneous values. The current in
the resistor is in phase with the applied
voltage, and the current in the capacitor
leads the applied voltage by 90°. The
net current phasor is the sum of the
r r r
branch current phasors I = I C + I R .
(
r
ε
r
IC
max
r
IR
δ
ωt
)
The peak current through the parallel
combination is equal to ε peak Z , where
Z is the impedance of the combination:
From the phasor diagram we have:
r
I
I = I peak cos(ωt − δ ) ,
where I peak =
ε peak
Z
2
= I R2 , peak + I C2 , peak
I peak
ε  ε 
=  peak  +  peak 
 R   XC 
2
2
2
 1
1  ε peak


=ε  2 + 2  = 2
XC  Z
R
1
1
1
where 2 = 2 + 2
Z
R
XC
2
peak
Solving for Ipeak yields:
From the phasor diagram:
I peak =
ε peak
Z
where Z −2 = R −2 + X C−2
I = I peak cos(ωt + δ )
where
tan δ =
ε peak
IC
R
X
= C =
I R ε peak
XC
R
80 •
A transformer has 400 turns in the primary and 8 turns in the secondary. (a) Is
this a step-up or a step-down transformer? (b) If the primary is connected to a 120 V rms
voltage source, what is the open-circuit rms voltage across the secondary? (c) If the
primary rms current is 0.100 A, what is the secondary rms current, assuming negligible
magnetization current and no power loss?
Picture the Problem Let the subscript 1 denote the primary and the subscript 2
the secondary. We can decide whether the transformer is a step-up or step-down
transformer by examining the ratio of the number of turns in the secondary to the
number of terms in the primary. We can relate the open-circuit rms voltage in the
secondary to the primary rms voltage and the turns ratio.
(a) Because there are fewer turns in the secondary than in the primary it is a stepdown transformer.
(b) Relate the open-circuit rms
voltage V2, rms in the secondary to the
rms voltage V1, rms in the primary:
V2, rms =
N2
V1, rms
N1
Substitute numerical values and
evaluate V2, rms :
V2, rms =
8
(120 V ) = 2.40 V
400
(c) Because there are no power
losses:
V1, rms I1, rms = V2, rms I 2, rms
and
I 2, rms =
Substitute numerical values and
evaluate I2, rms:
I 2, rms =
V1, rms
V2, rms
I1, rms
120 V
(0.100 A ) = 5.00 A
2.40 V