1. In a Half-wave controlled rectifier calculate the average value of the voltage if the supply is 13sin(25t) and the firing angle is 13°.
A. 4.08 V
B. 4.15 V
C. 3.46 V
D. 5.48 V
Answer: A
In Half-wave controlled rectifier, the average value of the voltage is
Vm(1+cos(∝))÷2π
=13(1+cos(13°))÷6.28=4.08 V.
The thyristor will conduct from ∝ to π.
2. Calculate the extinction angle as a purely inductive load if the firing angle is 13°.
A. 328°
B. 347°
C. 349°
D. 315°
Answer: B
The extinction angle in the purely inductive load is
2π-∝=360°-13°=347°.
The extinction angle is the angle at which the current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.
3. Calculate the conduction angle as a purely inductive load if the firing angle is 165°.
A. 78°
B. 55°
C. 30°
D. 19°
Answer: C
The conduction angle in the purely inductive load is
β-α=2(π-∝)
=2(180°-165°)=30°.
The conduction angle is the angle at which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.
4. R-L-C underdamped loads are generally lagging power factor loads.
A. True
B. False
Answer: B
R-L-C underdamped loads are generally leading power factor loads. They do not require forced commutation. Anti-Parallel diodes help in the commutation process.
5. In a Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 3sin(5t).
A. .95 V
B. .92 V
C. .93 V
D. .94 V
Answer: A
In a Half-wave uncontrolled rectifier, the average value of the voltage is
Vm÷π=3÷π=.95 V.
The diode will conduct only for the positive half cycle. The conduction period of the diode is π.
6. In a Half-wave uncontrolled rectifier calculate the r.m.s value of the voltage if the supply is 89sin(41t). [/bg_collapse]
A. 91.5 V
B. 44.5 V
C. 25.1 V
D. 15.1 V
Answer: B
In a Half-wave uncontrolled rectifier, the r.m.s value of the voltage is
Vm÷2
=89÷2=44.5 V.
The diode will conduct only for the positive half cycle. The conduction period of the diode is π.
7. In a Half-wave uncontrolled rectifier calculate the power dissipation across the 8 Ω resistor if the supply is 29sin(22t).
A. 26.2 W
B. 24.2 W
C. 26.1 W
D. 29.1 W
Answer: A
In a Half-wave uncontrolled rectifier, the r.m.s value of the voltage is
Vm÷2=29÷2=14.5 V.
The diode will conduct only for the positive half cycle. Power dissipation across the resistor is
V2r.m.s÷R
=14.52÷8=26.2 W.
8. The conduction period of diode in Half-wave uncontrolled rectifier for resistive load is ______________
A. π
B. 2π
C. 3π
D. 4π
Answer: A
The conduction period of the diode in a Half-wave uncontrolled rectifier for the resistive load is π. For the negative A.C supply diode will be reverse biased.
9. In a Half-wave uncontrolled rectifier calculate the average value of the current for 3 Ω resistive load if the supply is 34sin(11t).
A. 3.6 A
B. 2.6 A
C. 2.5 A
D. 3.1 A
Answer: A
In a Half-wave uncontrolled rectifier, the average value of the current is
Vm÷πR
=34÷3π=3.6 A.
The diode will conduct only for the positive half cycle. The conduction period of the diode is π.
10. In a Half-wave controlled rectifier calculate the average value of the current for 2.5 Ω resistive load if the supply is sin(5.2t) and the firing angle is 26°.
A. 0.8 V
B. 0.15 V
C. 0.12 V
D. 0.21 V
Answer: C
In Half-wave controlled rectifier, the average value of the current is