Electric Drives Questions and Answers – Solid-State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Inductive Loads Without Freewheeling Diode

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This set of Electric Drives online test focuses on “State Switching Circuits – Single Phase, Half-Wave, AC/DC Conversion for Inductive Loads Without Freewheeling Diode”.

1. Calculate the value of the conduction angle for R-L load if the value of β and α are 19° and 29°.
a) 10°
b) 70°
c) 30°
d) 80°
View Answer

Answer: a
Explanation: The conduction angle for R-L load is β-α=29°-19°=10°. R-L load is a current stiff type of load. The current in the circuit only flows from α to β.
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2. Calculate the V0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=24 V, ∝=30°, β=60°.
a) 1.39 V
b) 8.45 V
c) 4.55 V
d) 1.48 V
View Answer

Answer: a
Explanation: In Half-wave controlled rectifier, the average value of the voltage is Vm(cos(∝)-cos(β))÷2π=24(cos(30°)- cos(60°))÷6.28=1.39 V. The thyristor will conduct from ∝ to β.

3. Calculate the I0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=56 V, ∝=15°, β=30°, R=2 Ω.
a) .56 A
b) .44 A
c) .26 A
d) .89 A
View Answer

Answer: a
Explanation: In Half-wave controlled rectifier, the average value of the current is Vm(cos(∝)-cos(β))÷2πR=56(cos(15°)- cos(30°))÷12.56=.44 A. The thyristor will conduct from ∝ to β.

4. Calculate the I0 r.m.s for Single-phase Half Wave rectifier for R-L load (Highly inductive) using the data: Vm=110 V, ∝=16°, β=31°, R=1 Ω.
a) 4.56 A
b) 1.82 A
c) 4.81 A
d) 9.15 A
View Answer

Answer: b
Explanation: In Half-wave controlled rectifier, the r.m.s value of the current is Vm(cos(∝) – cos(β))÷2πR=110(cos(16°) – cos(31°))÷6.28=1.82 A. The thyristor will conduct from ∝ to β.

5. Calculate the value of the fundamental displacement factor for 1-φ Full wave semi-converter if the firing angle value is 0o.
a) 1
b) .8
c) .4
d) .2
View Answer

Answer: a
Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-φ Full wave semi-converter is cos(∝÷2)=cos(0o)=1.
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6. For highly inductive load current remains continuous.
a) True
b) False
View Answer

Answer: a
Explanation: For highly inductive load current remains continuous. Inductor opposes the change in the current. When current is about to extinct the next conducting cycle starts due to which currently remains constant.

7. Calculate the value of the Input power factor using the data: Fundamental displacement factor=.96, Distortion factor=.97.
a) .93
b) .84
c) .48
d) .89
View Answer

Answer: a
Explanation: The value of the input power factor is a product of the (Fundamental displacement factor)×(Distortion factor)=g×F.D.F=.97×.96=.93.

8. Displacement factor depends upon the shape of the waveform.
a) True
b) False
View Answer

Answer: a
Explanation: Displacement factor depends upon the shape of the waveform. Displacement factor is the cosine of the angle between the fundamental voltage and fundamental current. It depends upon the type of load, converter type.

9. In single phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (Vs)r.m.s=340 V, f=50 Hz, R=2 Ω, E=150 V, β=160°.
a) 45.89 V
b) 74.45 V
c) 54.85 V
d) 84.48 V
View Answer

Answer: b
Explanation: In single phase RLE load the voltage across the thyristor when current decays to zero VT=Vmsin(β)-E=340×√2sin(160°)-90=74.45 V.
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10. In single phase RLE load, calculate the angle at which conduction starts using the data: (Vs)=14sin(Ωt), f=50 Hz, E=10 V.
a) 45.58°
b) 46.26°
c) 47.26°
d) 49.56°
View Answer

Answer: a
Explanation: The angle at which conduction starts when Vmsin(θ)=E. The conduction will remain from ϴ to π-θ. Θ=sin-(E÷Vm)=sin-(.71)=45.58°.

11. Full form of PIV.
a) Peak inverse voltage
b) Peak insert voltage
c) Paas Inverse volatile
d) Peak insert volatile
View Answer

Answer: a
Explanation: PIV stands for Peak inverse voltage. It is the maximum negative voltage after which thyristor will breakdown.

12. Calculate the PIV for the Full-wave bridge rectifier if the peak value of the supply voltage is 230.
a) 324.4 V
b) 325.2 V
c) 524.4 V
d) 626.8 V
View Answer

Answer: a
Explanation: The peak inverse voltage for the Full-wave bridge rectifier is Vm=√2×230=325.2 V. The peak inverse is the maximum negative voltage across the thyristor.

13. In single phase RLE load, calculate the Peak inverse voltage using the data: (Vs)r.m.s=89 V, f=50 Hz, E=440 V.
a) 521.2 V
b) 527.8 V
c) 597 V
d) 529 V
View Answer

Answer: c
Explanation: In a single phase, RLE load the peak inverse voltage across the thyristor VT=Vm+E=529 V.
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14. Full form of DAC is ___________
a) Digital to Analog converter
b) Discrete to Analog converter
c) Digital to AC converter
d) Discrete to Avalanche converter
View Answer

Answer: a
Explanation: DAC stands for Digital to Analog converter. It converts the digital signal to an analog signal. It is used in microcontrollers and microprocessors.

15. Full form of ADC is ___________
a) Analog to Digital converter
b) Discrete to Analog converter
c) Analog to Digital converter
d) Discrete to Avalanche converter
View Answer

Answer: c
Explanation: ADC stands for Analog to Discrete converter. It converts the analog signal to a discrete signal. It is used in microcontrollers and microprocessors.

Sanfoundry Global Education & Learning Series – Electric Drives.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn