Why the efficiency is same in both questions of single phase half wave the first is R load and the second is R-L load?
Why the efficiency is same in both questions of single phase half wave the first is R load and the second is R-L load?
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Why the efficiency is same in both questions of single phase half wave the first is R load and the second is R-L load?
Transcribed Image Text:Example 1: For the shown half-wave rectifier, the voltage source is a 120 V with a
frequency of 60 Hz. The load resistor is 5 2. Determine:
(a) The average load current.
(b) The de and ac power absorbed by the load.
(c) The input power factor of the circuit.
(d) The efficiency of the rectifier.
Solution:
VS.RMS 120 V, F = 60 Hz, R = 50
(a)
(b)
(c)
(d)
=
Inc(load) =
Inc(load) =
Vm
TR
120√2
5π
Ppc = Vpc(load) IDC(load)
PDc = (54) (10.8) = 583.2 Watt
PF=
= 10.8 A
PAC = VRMS (load) IRMS(load)
PAC (84.9) (17) = 1443.3 Watt
=
PAC(load)
VS,RMS IS,RMS
n=VSRMS ¹S,RMS
=
1443.3
(120)(17)
PDC (load)* 100% =
}
+
Vm = √2 VS.RMS
V₂ = 120 V
F = 60 Hz
VDC(load) =
VRMS(load) =
= 0.707
IRMS(load) =
583.2
(120)(17)
Vm
TT
Vm
2
=
=
VRMS
R
120√2
120√2
2
=
It
*100% 28.5%
84.9
5
592
Page | 18
= 54 V
= 84.9 V
= 17 A
Page | 19
![Solution:
Example 2: For the half-wave uncontrolled rectifier with R-L load, the voltage source
is a 70.7 V with a frequency of 60 Hz. The load comprise resistance of 100 2 and
inductance of 0.1 H. Assume the conduction angle equal to 3.5 rad and then determine:
VERMS 70.7 V,
R = 100 2.
(a) An expression for the current in this circuit.
(b) The average current and voltage.
(c) The mms voltage.
(d) The efficiency of the rectifier if the RMS current is equal to 0.47 A.
F = 60 Hz, w = 377 rad/sec,
(b)
(c)
(d)
i(wt) =
tan
T= R
[31] .....
((wt) =
=
locitoad) =
n=
V = 100 V,
L= 0.1H, B = 3.5 rad 201⁰
Z= √R² + (wl.)² = √(100)³ + (37.7)² = 106.92
-² (1²)
L 0.1
100 = 0,001
n =
Vncia)=
VRMs(load)
sin(wt - 0) + sin(0)e#.
V₁
[pe(o)=[1-cas(f)]
Vestload) =
V₂
Voccoud) = [1-cas(8)]
2π
0
= tan-1
100
2π
-¹ (300) =
Puc
VARMS SMS
0.936 sin(wt - 20.7) + sin(20.7) 7
100
2m(100) [1-cos(201)] = 0.308 A
•
411
100=20,7²
20.7⁰
[1 − cos(201)] = 30.8 V
(²8-sin(28)
0
100% =
COO
*
(30.8)(0.308)
* 100% = 28.5%
(70.7) (0.47)
0≤ wt≤p
B<wt < 2x
(100) 3.5-sin(2(201)) = 50.2 V
4π
Vacp) 100%
VERMS M
F
P
6:37 ص
Page 23
0swtsB
B≤w ≤ 21
Page 24](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3c6eeaea-dc38-4396-9882-98b1d066bc2e%2F3722b905-55c2-482e-b12c-b5cf1a592f35%2F2oyocu9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solution:
Example 2: For the half-wave uncontrolled rectifier with R-L load, the voltage source
is a 70.7 V with a frequency of 60 Hz. The load comprise resistance of 100 2 and
inductance of 0.1 H. Assume the conduction angle equal to 3.5 rad and then determine:
VERMS 70.7 V,
R = 100 2.
(a) An expression for the current in this circuit.
(b) The average current and voltage.
(c) The mms voltage.
(d) The efficiency of the rectifier if the RMS current is equal to 0.47 A.
F = 60 Hz, w = 377 rad/sec,
(b)
(c)
(d)
i(wt) =
tan
T= R
[31] .....
((wt) =
=
locitoad) =
n=
V = 100 V,
L= 0.1H, B = 3.5 rad 201⁰
Z= √R² + (wl.)² = √(100)³ + (37.7)² = 106.92
-² (1²)
L 0.1
100 = 0,001
n =
Vncia)=
VRMs(load)
sin(wt - 0) + sin(0)e#.
V₁
[pe(o)=[1-cas(f)]
Vestload) =
V₂
Voccoud) = [1-cas(8)]
2π
0
= tan-1
100
2π
-¹ (300) =
Puc
VARMS SMS
0.936 sin(wt - 20.7) + sin(20.7) 7
100
2m(100) [1-cos(201)] = 0.308 A
•
411
100=20,7²
20.7⁰
[1 − cos(201)] = 30.8 V
(²8-sin(28)
0
100% =
COO
*
(30.8)(0.308)
* 100% = 28.5%
(70.7) (0.47)
0≤ wt≤p
B<wt < 2x
(100) 3.5-sin(2(201)) = 50.2 V
4π
Vacp) 100%
VERMS M
F
P
6:37 ص
Page 23
0swtsB
B≤w ≤ 21
Page 24
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