1. The single-phase half-controlled rectifier shown in Fig.1 operating at a triggering angle of a = 60° from A.C. source Vs = 300 sinot. Assuming the load is RL, where, R = 10 2, L= 20 mH, at 60 Hz. Assume the distinction angle ß = 185 degree. 2. Determine the following: A. express Vde of the load as a function of a, and calculate its value. B. Determine an expression for the current in the circuit. C. The power absorbed by the resistor.
1. The single-phase half-controlled rectifier shown in Fig.1 operating at a triggering angle of a = 60° from A.C. source Vs = 300 sinot. Assuming the load is RL, where, R = 10 2, L= 20 mH, at 60 Hz. Assume the distinction angle ß = 185 degree. 2. Determine the following: A. express Vde of the load as a function of a, and calculate its value. B. Determine an expression for the current in the circuit. C. The power absorbed by the resistor.
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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![1. The single-phase half-controlled rectifier shown in
Fig.1 operating at a triggering angle of a = 60°
from A.C. source Vs = 300 sinot. Assuming the
load is RL, where, R = 10 2, L = 20 mH, at 60 Hz.
Assume the distinction angle ß = 185 degree.
2.
Determine the following:
A. express Vde of the load as a function of a, and
calculate its value.
B. Determine an expression for the current in the
circuit.
C. The power absorbed by the resistor.
IL
1:1
VL
Vp = Vnsinot
D
ell](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F55abc52f-d869-4deb-bd7d-8499ac8469f2%2F1086212b-6e74-4604-9b15-c4729d18f631%2Fzxy1dhd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. The single-phase half-controlled rectifier shown in
Fig.1 operating at a triggering angle of a = 60°
from A.C. source Vs = 300 sinot. Assuming the
load is RL, where, R = 10 2, L = 20 mH, at 60 Hz.
Assume the distinction angle ß = 185 degree.
2.
Determine the following:
A. express Vde of the load as a function of a, and
calculate its value.
B. Determine an expression for the current in the
circuit.
C. The power absorbed by the resistor.
IL
1:1
VL
Vp = Vnsinot
D
ell
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