Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 2.
Chapter 10, Problem 6.
(a) If -10 cos ωt + 4 sin ωt = A cos (ωt + φ), where A > 0 and -180° < φ ≤ 180°, find A and φ.
(b) If 200 cos (5t +130°) = F cos 5t + G sin 5t, find F and G.
(c) Find three values of t, 0 ≤ t ≤ 1 s, at which i(t) = 5 cos 10t - 3 sin 10t = 0.
(d) In what time interval between t = 0 and t = 10 ms is 10 cos 100πt ≥ 12 sin 100πt?
Compare the following pairs of wave forms, and determine which one is leading:
(a) -33 sin (8t - 9°) and 12 cos (8t - 1°)
(b) 15 cos (1000t + 66°) and -2 cos (1000t + 450°).
(c) sin (t - 13°) and cos (t - 90°).
(d) sin t and cos (t -90°).
Chapter 10, Solution 2.
Chapter 10, Solution 6.
(a)
−10 cos ωt + 4 sin ωt +=A
ACos ( wt + Φ ), A > 0, − 180° < Φ ≤ 180°
(a)
A = 116 = 10.770, A cos Φ = −10, A sin Φ = −4 ∴ tan Φ = 0.4, 3d quad
-33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o
12 cos (8t – 1o) → 12∠-1o
33∠81o
∴Φ = 21.80° = 201.8°, too large ∴Φ = 201.8° − 360° = −158.20°
(b)
(c)
(d)
-33 sin(8t – 9o) leads 12 cos (8t – 1o) by
81 – (-1) = 82o.
200 cos (5t + 130°) = Fcos 5t + G sin 5t ∴ F = 200cos130° = −128.56
G = −200sin130° = −153.21°
sin10t 5
= , 10t = 1.0304,
i(t ) = 5cos10t − 3sin10t = 0, 0 ≤ t ≤ 1s ∴
cos10t 3
t = 0.10304 s; also, 10t = 1.0304 + π, t = 0.4172s; 2π : 0.7314s
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12∠-1o
(b)
15 cos (1000t + 66o)
-2 cos (1000t + 450o)
0 < t < 10ms, 10 cos100πt ≥ 12sin100πt ; let 10cos100πt =12sin100πt
10
∴ tan100πt = , 100πt = 0.6947 ∴ t = 2.211ms ∴ 0 < t < 2.211ms
12
→
→
15 ∠ 66o
-2 ∠ 450o = -2 ∠90o = 2 ∠ 270o
15 cos (1000t + 66o) leads -2 cos (1000t + 450o) by 66 – -90 =
156o.
15∠66o
2∠270o
(c)
sin (t – 13o)
cos (t – 90o)
→
→
1∠-103o
1 ∠ -90o
cos (t – 90o) leads sin (t – 13o)
by 66 – -90 = 156o.
1∠-103o
(d)
1 ∠ -90o
sin t
cos (t – 90o)
→
→
1 ∠ -90o
1 ∠ -90o
These two waveforms are in phase. Neither leads the other.
Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 30.
Chapter 10, Problem 18.
Assume that the op-amp in Fig. 10.50 is ideal (Ri = ∞ , Ro = 0, and A = ∞ ).
Note also that the integrator input has two signals applied to it, -Vm cos ωt and vout. If the product
R1C1 is set equal to the ratio L/R in the circuit of Fig. 10.4, show that vout equals the voltage
across R in Fig. 10.4.
Let ω= 4 krad/s, and determine the instantaneous value of ix at t = 1ms if I x equals:
(a) 5∠ − 80° A
(b) −4 + j1.5 A
Express in polar form the phasor voltage Vx if vx (t ) equals:
(c) 50 sin (250t - 40°)
(d) 20 cos108t − 30sin108t V
(e) 33cos (80t − 50°) + 41cos (80t − 75°) V
Chapter 10, Solution 30.
ω = 4000, t = 1ms
(a)
and ω t = 4 rad
I x = 5∠ − 80° A
∴ ix = 54 cos (4rad − 80°) = −4.294 A
Chapter 10, Solution 18.
(b)
L
R i = ∞, R o = 0, A = ∞, ideal, R1C1 =
R
v
V cos ω t
iupper = − m
, ilower = out
R
R1
∴ ic1 = iupper + ilower =
i
′
(vout − Vm cos ω t ) = −C1vout
R1
L
′
vout
R
d v 
For RL circuit, Vm cos ω t = vr + L  R 
dt  R 
L
∴ Vm cos ω t = vR + vR′
R
By comparison, vR = vout
I x = −4 + j1.5 = 4.272∠159.44° A
∴ ix = 4.272 cos (4rad + 159.44°) = 3.750− A
(c)
vx (t ) = 50sin (250t − 40°)
= 50 cos (250t − 130°) → Vx = 50∠ − 103° V
(d)
′ = vout +
∴ Vm cos ω t = vout + R1C1vout
vx = 20 cos108t − 30sin108t
→ 20 + j 30 = 36.06∠56.31° V
(e)
vx = 33cos (80t − 50°) + 41cos (80t − 75°)!
→ 33∠ − 50° + 41∠ − 75° = 72.27∠ − 63.87° V
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 32.
Chapter 10, Problem 34.
If ω=500 rad/s and IL=2.5 ∠40° A in the circuit of Fig. 10.53, find vs(t).
A is flowing through the series combination of 1 Ω,
A phasor current of 1
1 H, and 1 F. At what frequency is the amplitude of the voltage across the
network twice the amplitude of the voltage across the resistor?
Chapter 10, Solution 34.
VR = 1∠0o V, Vseries = (1 + jω –j/ω)(1∠0o)
VR = 1 and Vseries = 1 + (ω - 1/ω )
2
We desire the frequency ω at which Vseries = 2VR or Vseries = 2
2
Thus, we need to solve the equation 1 + (ω - 1/ω ) = 4
Chapter 10, Solution 32.
or ω 2 - 3 ω - 1 = 0
Begin with the inductor: ZL = jωL = j 500 rad/s 0.02 Vs/A = j10 Ω
o
-3
o
(2.5 ∠40 ) (j500) (20×10 ) = 25∠130 V across the inductor and the 25-Ω resistor.
The current through the 25-Ω resistor is then (25∠130o) / 25 = 1∠130o A.
The current through the unknown element is therefore 2.5∠40o + 1∠130o = 2.69∠61.80o;
this is the same current through the 10-Ω resistor as well. Thus, KVL provides that
Vs = 10(2.69∠61.80o) + (25 ∠ -30o) + (25∠130o) = 35.44 ∠58.93o
and so vs(t) = 35.44 cos (500t + 58.93o) V.
Solving, we find that ω = 2.189 rad/s.
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 37.
Find Zin at terminals a and b in Fig. 10.56 if ω equals (a) 800 rad/s; (b) 1600 rad/s.
Chapter 10, Solution 37.
(a)
(b)
ω = 800 : 2µF → − j 625, 0.6H → j 480
300(− j 625) 600( j 480)
∴ Zin =
+
300 − j 625 600 + j 480
= 478.0 + j175.65Ω
300(− j312.5)
ω = 1600 : Zin =
300 − j312.5
600( j 960)
+
= 587.6 + j119.79Ω
600 + j 960
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 39.
If a voltage source vs = 120 cos 800t V is connected to terminals a and b in Fig. 10.56 (+
reference at the top), what current flows to the right in the 300-Ω resistance?
Chapter 10, Solution 39.
ω = 800 : 2µF → − j 625, 0.6H → j 480
300(− j 625) 600( j 480)
∴ Zin =
+
300 − j 625 600 + j 480
= 478.0 + j175.65Ω
120
− j 625
×
478.0 + j175.65 300 − j 625
or I = 0.2124 ∠ − 45.82° A
∴I =
Thus, i(t) = 212.4 cos (800t – 45.82o) mA.
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
Chapter 10, Problem 50.
Chapter 10, Problem 54.
Two admittances, Y1 = 3 +j4 mS and Y2 = 5 + j2 mS, are in parallel, and a third admittance, Y3 =
2 - j4 mS, is in series with the parallel combination.
Use phasors and mesh analysis on the circuit of Fig. 10.63 to find IB .
If a current I1 = 0.1
A is flowing through Y1, find the magnitude of the voltage across (a)
Y1; (b) Y2; (c) Y3; (d) the entire network.
Chapter 10, Solution 50.
I1
0.1∠30°
=
= 20∠ − 23.13°∴ V1 = 20 V
Y1 (3 + j 4)10−3
(a)
V1 =
(b)
V2 = V1 ∴ V2 = 20V
(c)
I 2 = Y2 V2 = (5 + j 2)10 −3 × 20∠ − 23.13° = 0.10770∠ − 1.3286° A
Chapter 10, Solution 54.
∴ I3 = I1 + I2 = 0.1∠30° + 0.10770∠ − 1.3286° = 0.2∠13.740° A
∴ V3 =
(d)
I3 0.2∠13.740°
=
= 44.72∠77.18° V ∴ V3 = 44.72V
Y3 (2 − j 4)10−3
Vin = V1 + V3 =+ 20∠ − 23.13° + 44.72∠77.18° = 45.60∠51.62°
∴ Vin = 45.60V
-2jI
j 3I B − j 5(I B − I D ) = 0 ∴−
2IBB + j 5I D = 0
3(I D + j 5) − j5(I D − I B ) + 6 (I D + 10) = 0
∴ j 5I B + (9 − j5) I D = −60 − j15
0
j5
−60 − j15 9 − j 5 −75 + j 300
IB =
=
j5
− j2
15 − j18
j5 9 − j5
= 13.198∠154.23° A
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Examples - Chapter 10 - Sinusoidal Steady-State Analysis
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Chapter 10, Problem 67.
Chapter 10, Problem 82.
Find the input admittance of the circuit shown in Fig. 10.74, and represent it as the parallel
combination of a resistance R and an inductance L , giving values for R and L if ù =1 rad/s.
In the circuit of Fig. 10.85, find values for (a) I1, I2, and I3.(b) Show Vs, I1, I2, and I3 on a phasor
diagram (scales of 50 V/in and 2 A/in work fine). (c) Find Is graphically and give its amplitude
and phase angle.
Chapter 10, Solution 67.
Note: There is no dependent but one independent source. Therefore, find the input
impedance/admittance through an experiment, .e.g., apply an input current of 1 ∠0o A.
Chapter 10, Solution 82.
, 0.5VL = jω
VL ==jω1∠0
j 210o∴
∴ Vin = (1 + jω )
1
+ j 2ω
jω
1
+ j 2ω
jω
V
1
∴ Zin = in = 1 +
+ j 2ω
jω
1
At ω = 1, Zin = 1 − j1 + j 2 = 1 + j
(a)
= 1+
∴ Yin =
1
= 0.5 + j 0.5
1 + j1
so Yin =
ω
ω + j (2ω 2 − 1)
120
= 3∠ − 30° A
40∠30°
120
= 2.058∠30.96° A
I2 =
50 − j 30
120
= 2.4∠ − 53.13° A
I3 =
30 + j 40
I1 =
(b)
R = 500 mΩ, L = 500 mH.
(c)
I s = I1 + I 2 + I3
= 6.265∠ − 22.14° A