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°.

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.

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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

V^{2}r.m.s÷R
=14.5^{2}÷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