SOLUTION: E=200V + Riz 40kr www Find fp Resonant freq fp: = fp CVC L=200 mH -0.01MF Q₁=20 L 211 200X10X0.01 x 10 = 3558.81 Hz b. Find Vc Wp - 21fp Wp * 2X X3558.81 Wp = 22.36 K/sec Next, pind xt = WPL -3 7 22 x 10 x 200 x10 λ= 4472 Next, Frid Xc Xc = 1 = 21 мс Xc - 4472.271 jxc = -j 4472.271 JXL = 4472.1 22.36 X 0.01 X/0 Q= WL = 20 = 22.36 X18³ X 200 X10° Ri R₁ RI 223.62 3 apply ка Ve- 200 1 VC 90R + ye -0 2236 +4321 3443-311 Ve 0.00 2797000 to 238.6+ja -j12-071 50.000 10005 VC [00000x + (-0.002221) + 0.0001240.00024 Von 0.005 0.00000017 Ve 18.24 Find P? remoore power pooter is wity Pc VI Vcx Vo RI 238. 220.6 +4478-1 Ve. 1248 P-V (1243) X 19824 691mats R 229.6 Frd addih, BW-? Bur fy 83350 91 BU (78.0382 Q2. A/ For the parallel resonance circuit, find: 1. The center frequency fo 2. The quality factor and bandwidth of the circuit. 3. The output voltage and current in cach clements. - 0005 2m^ 15kn SMH 10nF

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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The expert solved the question correctly and I am grateful to him, but can he explain the solution in simple steps with the solution to the question sent at the end?
SOLUTION:
E=200V
+
Riz 40kr
www
Find fp
Resonant freq
fp:
=
fp
CVC
L=200 mH
-0.01MF
Q₁=20
L
211 200X10X0.01 x 10
= 3558.81 Hz
b. Find Vc
Wp - 21fp
Wp * 2X X3558.81
Wp = 22.36 K/sec
Next, pind xt = WPL
-3
7 22 x 10 x 200 x10
λ= 4472
Next, Frid Xc
Xc = 1 =
21 мс
Xc - 4472.271
jxc = -j 4472.271
JXL = 4472.1
22.36 X 0.01 X/0
Q= WL = 20 = 22.36 X18³ X 200 X10°
Ri
R₁
RI 223.62
3
Transcribed Image Text:SOLUTION: E=200V + Riz 40kr www Find fp Resonant freq fp: = fp CVC L=200 mH -0.01MF Q₁=20 L 211 200X10X0.01 x 10 = 3558.81 Hz b. Find Vc Wp - 21fp Wp * 2X X3558.81 Wp = 22.36 K/sec Next, pind xt = WPL -3 7 22 x 10 x 200 x10 λ= 4472 Next, Frid Xc Xc = 1 = 21 мс Xc - 4472.271 jxc = -j 4472.271 JXL = 4472.1 22.36 X 0.01 X/0 Q= WL = 20 = 22.36 X18³ X 200 X10° Ri R₁ RI 223.62 3
apply ка
Ve- 200
1 VC
90R
+ ye
-0
2236 +4321
3443-311
Ve
0.00
2797000
to
238.6+ja -j12-071
50.000
10005
VC [00000x + (-0.002221) + 0.0001240.00024
Von 0.005
0.00000017
Ve 18.24
Find P?
remoore power pooter is wity
Pc VI
Vcx Vo
RI
238.
220.6 +4478-1
Ve. 1248
P-V (1243)
X 19824
691mats
R
229.6
Frd addih, BW-?
Bur fy
83350 91
BU (78.0382
Q2. A/ For the parallel resonance circuit, find:
1. The center frequency fo
2. The quality factor and bandwidth of the circuit.
3. The output voltage and current in cach clements.
-
0005
2m^
15kn
SMH
10nF
Transcribed Image Text:apply ка Ve- 200 1 VC 90R + ye -0 2236 +4321 3443-311 Ve 0.00 2797000 to 238.6+ja -j12-071 50.000 10005 VC [00000x + (-0.002221) + 0.0001240.00024 Von 0.005 0.00000017 Ve 18.24 Find P? remoore power pooter is wity Pc VI Vcx Vo RI 238. 220.6 +4478-1 Ve. 1248 P-V (1243) X 19824 691mats R 229.6 Frd addih, BW-? Bur fy 83350 91 BU (78.0382 Q2. A/ For the parallel resonance circuit, find: 1. The center frequency fo 2. The quality factor and bandwidth of the circuit. 3. The output voltage and current in cach clements. - 0005 2m^ 15kn SMH 10nF
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