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DFCCIL’21 EE: Daily Practices Quiz 24-Aug-2021

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               DFCCIL’21 EE: Daily Practices Quiz 24-Aug-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06

Time : 8 min

Q1. For a transistor, collector current is 100 times the base current. Find β_dc.
(a) 100
(b) 90
(c) 10
(d) 50

Q2. An AC voltage of maximum value equal to 100V is applied to a single-phase fully controlled bridge circuit. The peak inverse voltage rating of each SCR used will be ______.
(a) 141.4 V
(b) 200 V
(c) 70.7 V
(d) 100 V

Q3. For a DC motor, gross mechanical power developed by the motor is maximum and the motor is supplied through a 220 V source. What will be back emf?
(a) 220 V
(b) 27.5 V
(c) 55 V
(d) 110 V

Q4. For a second order system having its transfer function as:
H(s)=25/(s^2+8s+25)
Find the damping factor.
(a) 0.4
(b) 0.8
(c) 0.5
(d) 0.6

Q5. KCL works on the principle of which of the following
(a) law of conservation of charge.
(b) law of conservation of energy.
(c) Both
(d) None of the above.

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SOLUTIONS

S1. Ans.(a)
Sol. ACQ: I_C=100I_B
⇒I_C/I_B =100=β_dc

S2. Ans.(d)
Sol. Peak inverse voltage (PIV): Peak inverse voltage is also referred to as reverse breakdown voltage or peak reverse voltage, which is defined as the maximum reverse voltage that a diode or PN-junction can withstand in a non-conducting state or reverse bias condition before breakdown.
PIV for single-phase fully controlled bridge circuit=V_m
ACQ: V_m=100 V
∴ PIV for single-phase fully controlled bridge circuit=V_m=100 V
NOTE: PIV for single phase mid-point full wave SCR converter=〖2V〗_m

S3. Ans.(d)
Sol. Condition for maximum power developed by the motor is: E_b=V/2
ACQ: V=220 V
∴E_b=V/2=220/2=110 V

S4. Ans.(b)
Sol. comparing with second order characteristics equation:
ω_n=5
and 2ξω_n=8
∴2ξ×5=8
⇒?=8/10=0.8

S5. Ans.(a)
Sol. KCL (Kirchhoff’s Current Law) states that in an electrical circuit total current entering in to a node is equal to the total current leaving from node. This works on the principle of law of conservation of charge. As the current is rate flow of charge. So, charge is conserved at a node.

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