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Q1: A rectifier is a:
A) Linear device.
B) Non-Linear device.
C) Bilateral device.
D) Passive device.
D) Passive device.
Q2: The use of capacitor filter in a rectifier circuit gives satisfactory performance only when the load:
A) Current is high
B) Current is low
C) Voltage is high
D) Voltage is low
Current is low
Q3: Transformer used in rectification to _________ the supply voltage.
A) Step up
B) Step down
C) equalize
D) None of these
B) Step down
Q4: Ripple factor of full wave rectifier?
A) 0.482
B) 1.21
C) 0.21
D) 0.84
A) 0.482
Q5: A full wave rectifier converts___________ into positive cycles.
A) Positive half cycle
B) Negative half cycle
C) Both positive and negative
D) None of these
C) Both positive and negative
Q6: Half wave rectifier convert _________ into positive half cycle.
A) Upper half cycle
B) Lower half cycle
C) Both half cycles
D) Either positive or negative half cycle
D) Either positive or negative half cycle
Q7: Ripple factor of half wave rectifier is:
A) 1.21
B) 0.486
C) 0.84
D) 0.5
A) 1.21
Q8: The half wave rectifier is equivalent to:
A) Clipper
B) Load
C) filter
D) Clamper
A) Clipper
Q9:A full wave rectifier uses load resistor of 1500Ω. Assume the diodes have Rf=10Ω, Rr=∞. The voltage applied to diode is 30V with a frequency of 50Hz. Calculate the AC power input:
A) 368.98 mW
B) 275.2 mW
C) 145.76 mW
D) 456.78 mW
B) 275.2 mW
The AC power input PIN=IRMS2(RF+Rr)
IRMS=Im/√2=Vm/(Rf+RL)√2
= 30/(1500+10)*1.414=13.5mA
So, PIN=(13.5*10-3)2*(1500+10)=275.2mW.
Q10: Efficiency of full wave rectifier is:
A) 81.6%
B) 41.5%
C) 50%
D) 100%
A) 81.2%
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