Ques 51. An op-amp has a common-mode gain of 0.01 and a differential mode gain of 105. Its common-mode rejection ratio would be
10-7
10-3
103
107
Answer.4. 107
Explanation:
CMMR = Differential mode gain / Common-mode gain = Ad/Ac
= 105/0.01
= 107
Ques 52. In which type of amplifier, the power output is highest possible?
Common C
Common B
Common E
Darlington Pair
Answer.4. Darlington Pair
Explanation:
Darlington pair consists of two transistors in such a way that the current amplified by the first transistor is amplified further by the second one. This results in a very high level of current gain. The overall current gain of the Darlington pair is the product of the two individual transistors.
This means that if two transistors with modest current gains of 25 were used, then the overall current gain would be 25 × 25 = 625
Ques 53. A voltage source VAB = 4 sin ωt is applied to the terminal A and B of the circuit shown in the given figure. The diodes are assumed to be ideal. The impedance offered by the circuit across the terminal A and B is
5 kΩ
10 kΩ
15 kΩ
20 kΩ
Answer.2. 10 KΩ
Explanation:
In the first half of the cycle, Diode D1 conducts and the impedance offered will be 10 KΩ as shown in the below figure
In the negative half cycle, diode D2 conducts and impedance offer will be 10 KΩ
Hence equivalent impedance offered across terminals A and B is 10 KΩ
Ques 54. Which of the following motor is best suitable for the hoist?
Crane and hoists require high starting torque during starting because they carry a heavy load. The induction motor has a starting torque of about 120 to 150 percent of full-load torque hence they are preferred for cranes and hoists.
Ques 55. The op-amp as shown in the figure has a gain of -100. The value of R1 is
10 KΩ
10.1 KΩ
20 KΩ
100 MΩ
Answer.1. 10 KΩ
Explanation:
OP-amp gain is given as
Gain = -(R2/R1)
-100 = -( 1 x 106/R1)
R1 = 1000Ω
R1 = 10 KΩ
Ques 56. Millman theorem is an extension of?
Norton
Thevenin
Superposition
Both 1 & 2
Answer.4. Both 1 & 2
Explanation:
Millman’s theorem provides an equivalent circuit that is Similar to Thevenin or Norton’s equivalent circuit. Hence we can say that Millman’s theorem is an extension of Thevenin’s or Norton’s theorem for the circuit containing the number of independent voltage or current sources respectively.
Ques 57. A tank circuit of RF amplifier has a coil of Q factor of 100. It is tuned to a frequency of 1 MHz. The bandwidth of the circuit is
1 kHz
10 kHz
100 kHz
1000 kHz
Answer.3. 10 KHz
Explanation:
Bandwidth = Frequency/Q Factor
= 1 x 106/100
=10 KHz
Ques 58. Diode resistance in case of forward biased in an ideal case
Very very Low
Zero
Infinite
Very High
Answer.2. Zero
Explanation:
In the ideal case diode resistance in forward biased acts as the perfect conductor and in reverse biased, it will act as the perfect insulator.
Therefore in an ideal case, the diode will offer zero resistance when forward biased.
Ques 59. If the resonant frequency of the circuit shown in figure I is 1 kHz, the resonant frequency of the circuit shown in figure II will be
4 kHz
2 KHz
0.05 KHz
0.25 KHz
Answer.3. 0.05 KHz
Explanation:
In figure second
Leq = 2L
Ceq = 2C
Ques 60. Two 100W, 200V lamps are connected in series across a 200V supply. The total power consumed by each lamp will be watts?
25 W
200 W
75 W
150 W
Answer.1. 25 W
Explanation:
We know that
P = V2/R
R = V2/P
R = (200*200)/100 = 400 ohms.
Since now the two bulbs are in series, the total resistance offered by them is :
R’ = 400 Ω + 400 Ω = 800 Ω
Now the current that flows through either of them is :
I = V/R
= 200/800 = 0.25A
Since the bulbs are in series, the same amount of current flows through each of them.
Thus the total power consumed by each lamp will be
