SSC CGL Tier I - 09 September 2024 - Shift 1
100 questions · 60 minutes · 4 subjects
SSC · Power Electronics and Drives practice
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English SSC CGL Tier I papers. Free 60-minute mocks, 100 questions, 200 marks; +2 correct and −0.5 incorrect. Source: SSC response sheets preserved by SSC Portal. The marked archived keys have not been independently revalidated against final SSC keys; worked explanations are pending review. A/B/C/D correspond to source options 1/2/3/4.
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Once triggered, an SCR continues to conduct until
Answer and explanation
Correct answer: B
The gate loses control after turn-on. The device turns off only when the anode current drops below the holding current.
In a single-phase half-wave uncontrolled rectifier with a resistive load, the average output voltage is
Answer and explanation
Correct answer: A
The average over a full cycle of a half-wave rectified sine is Vm/π. For a peak of 314 V this is about 100 V.
A step-down chopper has an input of 200 V and a duty ratio of 0.4. The average output voltage is
Answer and explanation
Correct answer: A
Vo = D × Vin = 0.4 × 200 = 80 V.
A cycloconverter converts
Answer and explanation
Correct answer: D
It changes the frequency without any intermediate dc link, and is used in large low-speed ac drives.
A power MOSFET is
Answer and explanation
Correct answer: B
The gate voltage controls the channel. It switches very fast, since there is no minority carrier storage.
The Ward-Leonard system is used for
Answer and explanation
Correct answer: D
A motor-generator set supplies a variable voltage to the armature of the dc motor, which is controlled through the generator field.
A single-phase fully controlled bridge rectifier has a peak supply voltage of 300 V and a firing angle of 60°. The average output voltage (continuous conduction) is about
Answer and explanation
Correct answer: A
Vdc = (2Vm/π) cos α = (600/π) × 0.5 = 95.5 V.
A step-up (boost) chopper has an input of 50 V and a duty ratio of 0.6. The output voltage is
Answer and explanation
Correct answer: C
Vo = Vin / (1 − D) = 50 / 0.4 = 125 V.
In a voltage source inverter, feedback diodes across the switches are needed to
Answer and explanation
Correct answer: B
With an inductive load, the current continues to flow when the switches turn off. The diodes return the energy to the dc source.
A TRIAC is equivalent to
Answer and explanation
Correct answer: D
It conducts in both directions and is triggered by a gate pulse of either polarity, so it is used for ac power control.
A snubber circuit across an SCR protects it against
Answer and explanation
Correct answer: B
The RC network slows the rise of the anode voltage, which would otherwise turn the device on falsely.
In the V/f control of an induction motor, the ratio of voltage to frequency is kept constant in order to
Answer and explanation
Correct answer: C
The flux is proportional to V/f. Constant flux gives a constant torque capability over the speed range below the base speed.
A step-up (boost) chopper has an input of 24 V and a duty ratio of 0.8. The output voltage is
Answer and explanation
Correct answer: C
Vo = Vin/(1 − D) = 24/0.2 = 120 V.
A single-phase half-wave controlled rectifier with a resistive load is fed from a supply of peak 141 V and fires at α = 30°. The average output voltage (Vm(1 + cosα)/2π) is about
Answer and explanation
Correct answer: D
Vdc = Vm(1 + cosα)/(2π) = 141 × (1 + cos 30°)/(2π) = 41.88 V.
An SCR in a circuit fed from a 255 V rms ac supply must withstand a maximum repetitive peak voltage. Allowing a safety factor of 2, the minimum voltage rating of the SCR is about
Answer and explanation
Correct answer: A
Peak = √2 × 255 = 360.6 V. With a factor of 2: 721 V.
An induction motor rated 415 V at 50 Hz is run at 30 Hz with constant V/f control. The stator voltage should be about
Answer and explanation
Correct answer: B
V/f = constant: V2 = 415 × 30/50 = 249 V.
A single-phase half-wave controlled rectifier with a resistive load is fed from a supply of peak 400 V and fires at α = 30°. The average output voltage (Vm(1 + cosα)/2π) is about
Answer and explanation
Correct answer: A
Vdc = Vm(1 + cosα)/(2π) = 400 × (1 + cos 30°)/(2π) = 118.79 V.
A single-phase half-bridge inverter has a dc supply of 360 V (each switch applies ±180 V). The rms output voltage is
Answer and explanation
Correct answer: A
The output is a square wave of amplitude ±180 V, so the rms value equals the amplitude 180 V.
A separately excited DC motor (Ra = 0.5 Ω) runs at 1200 rpm on 220 V with an armature current of 50 A. By armature voltage control the supply is reduced to 150 V (same flux and current). The new speed is about
Answer and explanation
Correct answer: C
Eb1 = 220 − 50 × 0.5 = 195 V; Eb2 = 150 − 50 × 0.5 = 125 V. N2 = N1 Eb2/Eb1 = 769 rpm.
A separately excited DC motor (Ra = 0.5 Ω) runs at 1000 rpm on 240 V with an armature current of 20 A. By armature voltage control the supply is reduced to 120 V (same flux and current). The new speed is about
Answer and explanation
Correct answer: B
Eb1 = 240 − 20 × 0.5 = 230 V; Eb2 = 120 − 20 × 0.5 = 110 V. N2 = N1 Eb2/Eb1 = 478 rpm.
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