

MCQOPTIONS
Saved Bookmarks
This section includes 380 Mcqs, each offering curated multiple-choice questions to sharpen your Analog Electronic Circuits knowledge and support exam preparation. Choose a topic below to get started.
1. |
The frequency of oscillation for a RC phase shift network is given by: |
A. | |
B. | |
Answer» D. | |
2. |
The frequency of oscillations in a Wien bridge oscillator given by: |
A. | |
B. | |
C. | |
Answer» B. | |
3. |
When FT is the transistors short-circuit commonemitter current gain-bandwidth, the maximum frequency of oscillations of a transistor will depend on: |
A. | (F |
B. | )2 |
C. | |
D. | |
E. | |
Answer» D. | |
4. |
In a Hartley oscillator the condition for sustained oscillations is
|
A. | |
B. | |
Answer» C. | |
5. |
The ideal characteristics of op-amp are:
|
A. | (i) and (ii) |
B. | (i), (ii) and (iii) |
C. | (i) and (iv) |
D. | All of these |
Answer» D. All of these | |
6. |
In the circuit shown below the input offset voltage and input offset current are Vio = 4 mV and Iio = 150 nA. The total output offset voltage is: |
A. | 479 mV |
B. | 234 mV |
C. | 168 mV |
D. | 116 mV |
Answer» B. 234 mV | |
7. |
In figure below assume that the diode and op-amps are ideal. For an input Vin = sin t, the output voltage V0 is: |
A. | full wave rectified with a peak value of 1 |
B. | full wave rectified with a peak value of 1 |
C. | half wave rectified with a peak value of 1 |
D. | half wave rectified with a peak value of 1 |
Answer» D. half wave rectified with a peak value of 1 | |
8. |
The output voltage vout of the given circuit, for a given input signal, as shown in the above Figure, will be of the form: |
A. | NA |
B. | NA |
C. | NA |
D. | NA |
Answer» D. NA | |
9. |
If Vi = 2 V, then output V0 is: |
A. | 6 V |
B. | 6 V |
C. | 3 V |
D. | 3 V |
Answer» E. | |
10. |
Calculate value of i0 |
A. | 18 A |
B. | 24 A |
C. | 24 A |
D. | 36 A |
Answer» C. 24 A | |
11. |
In a Schmitt trigger the output vo is limited by Zener diodes. Single Vs is connected to the negative input terminal (v ) and v+ = vo. The Hysteresis voltage is 1 volt. Triggering takes place for signals going through: |
A. | zero volt |
B. | 0 volt and + 1 volt |
C. | 0 volt and 1 volt |
D. | + 0.5 volt and 0.5 volt |
Answer» E. | |
12. |
If Vi = 2 V, then output V0 is: |
A. | 3 V |
B. | 6 V |
C. | 6 V |
D. | 3 V |
Answer» C. 6 V | |
13. |
The LED in the circuit shown below will be on if i is: |
A. | > 10 V |
B. | < 10 V |
C. | > 5 V |
D. | < 5 V |
Answer» D. < 5 V | |
14. |
In the circuit shown, R = 1 M (C = 0.5 F). Vs is a step voltage of magnitude 0.5 volt, V0 is given by: |
A. | + 0.5 volt |
B. | 0.5 volt |
C. | + t volt |
D. | t volt |
Answer» D. t volt | |
15. |
The Hysteresis phenomenon in Schmitt trigger has the following features:
|
A. | 2 and 3 only |
B. | 2, 3 and 5 only |
C. | 2, 3 and 4 only |
D. | 3 and 5 only |
Answer» C. 2, 3 and 4 only | |
16. |
The phase shift oscillator shown below operate at f = 80 kHz. The value of resistance RF is: |
A. | 148 k |
B. | 236 k |
C. | 438 k |
D. | 814 k |
Answer» C. 438 k | |
17. |
For a given circuit, the value of Vo and ACL is: |
A. | 5 V and 1 |
B. | 5 V and 5 |
C. | 0 V and 1 |
D. | 0 V and 5 |
Answer» B. 5 V and 5 | |
18. |
The maximum allowable value of Vin for circuit given below. Given gain of the amplifier is 200 |
A. | 400 mV |
B. | 40 mV |
C. | 4 V |
D. | 4000 mV |
Answer» C. 4 V | |
19. |
In the circuit shown below the op-amp is ideal. If F = 60, then the total current supplied by the 15 V source is: |
A. | 123.1 mA |
B. | 98.3 mA |
C. | 49.4 mA |
D. | 168 mA |
Answer» D. 168 mA | |
20. |
The value of C required for sinusoidal oscillation of frequency 1 kHz in the circuit shown below: |
A. | |
B. | |
C. | 2 F |
D. | 2 |
E. | F |
Answer» B. | |
21. |
Calculate the value of 0: |
A. | 6 V |
B. | 6 V |
C. | 10 V |
D. | 10 V |
Answer» D. 10 V | |
22. |
For the circuit given below V0: |
A. | V |
B. | |
C. | V |
D. | |
E. | |
Answer» C. V | |
23. |
In the op-amp circuit given below the load current iL is |
A. | |
B. | V |
C. | R |
Answer» B. V | |
24. |
Figure shown a Schmitt trigger circuit op-amp. The output V0 is limited to + 10 V and 5 V connecting suitably chosen Zener diodes at the output. The lower and upper trigger voltage are respectively: |
A. | 1 V and 0.5 V |
B. | 0.5 V and 1.0 V |
C. | 1 V and 0.5 V |
D. | 0.5 V and 1.0 V |
Answer» B. 0.5 V and 1.0 V | |
25. |
|
||||
A. | 8 | ||||
B. | 6 | ||||
C. | 6 | ||||
D. | 8 | ||||
Answer» D. 8 | |||||
26. |
Calculate V0 |
A. | 7.5 V |
B. | 7.5 V |
C. | 8 V |
D. | 8 V |
Answer» C. 8 V | |
27. |
Find V0 |
A. | 4 V |
B. | 4 V |
C. | 5 V |
D. | 5 V |
Answer» B. 4 V | |
28. |
In the circuit shown below | V0 | = 1 V to a certain set of values of , R and C. | V0 | will remain as 1 V even if: |
A. | None of these |
B. | is halved |
C. | R is halved |
D. | C is doubled |
Answer» B. is halved | |
29. |
In regard to improving the performance of a BJT differential amplifier, consider the following four statements:
|
A. | only 1, 2 |
B. | only 2, 3 |
C. | only 3, 4 |
D. | only 2, 4 |
Answer» E. | |
30. |
The bandwidth of an n-stage tuned amplifier with each stage having a bandwidth of B, is given by |
A. | |
B. | B |
Answer» D. | |
31. |
A differential amplifier has input v1 = 1050 V and v2 = 950 V, CMRR = 1000. The error in the differential output is: |
A. | 10% |
B. | 1% |
C. | 0.1% |
D. | 0.01% |
Answer» C. 0.1% | |
32. |
The following five statements are made with reference to the cascade amplifier:
|
A. | only 1, 3 |
B. | only 2, 4 |
C. | only 1, 3, 5 |
D. | only 2, 4, 5 |
Answer» B. only 2, 4 | |
33. |
In the circuit given below, Rf provides: |
A. | current series feedback |
B. | current shunt feedback |
C. | voltage series feedback |
D. | voltage shunt feedback |
Answer» E. | |
34. |
The slew rate of an operational amplifier is 0.5 V/micro sec. The maximum frequency of a sinusoidal input of 2 Vrms that can be handled without excessive distortion is: |
A. | 3 kHz |
B. | 30 kHz |
C. | 300 kHz |
D. | 2 MHz |
Answer» C. 300 kHz | |
35. |
Each transistor in the darlington pair shown below has hFE = 100. The overall hFE of the composite transistor neglecting leakage current is: |
A. | 10000 |
B. | 10001 |
C. | 10100 |
D. | 10200 |
Answer» E. | |
36. |
A differential amplifier has RL = 10 K (equal values in both collectors), hie = 1 K, Re = 50 K, hfe = 100. The common mode gain is given by: |
A. | 500 |
B. | 2 |
C. | 0.2 |
D. | 0.1 |
Answer» E. | |
37. |
Increasing the value of RE in a differential amplifier: |
A. | increases the differential gain A |
B. | increases the common mode gain A |
C. | increases CMRR |
D. | decreases CMRR |
Answer» D. decreases CMRR | |
38. |
In the op amp circuit, V0 is given by |
A. | 2 V |
B. | 8 V |
C. | |
D. | 8 V |
E. | 9 V |
Answer» E. 9 V | |
39. |
For the op-amp shown in fig below open loop differential gain is Aod = 103. The output voltage 0 for i = 2 V is: |
A. | 1.996 |
B. | 1.998 |
C. | 2.004 |
D. | 2.006 |
Answer» B. 1.998 | |
40. |
In the circuit, shown below I0 is given by: |
A. | |
Answer» C. | |
41. |
A circuit shown below. The largest value of RL that can be used is: |
A. | 100 |
B. | 400 |
C. | 2 k |
D. | 20 k |
Answer» C. 2 k | |
42. |
For the op-amp circuit shown below the voltage gain A = 0/ i is: |
A. | 8 |
B. | 8 |
C. | 10 |
D. | 10 |
Answer» B. 8 | |
43. |
For the circuit shown below gain is A = 0/ i = 10. The value of R is: |
A. | 600 k |
B. | 450 k |
C. | 4.5 M |
D. | 6 M |
Answer» C. 4.5 M | |
44. |
An op-amp has an open-loop gain of 105 and an openloop upper cut off frequency of 10 Hz. If this op-amp is connected as an amplifier with a closed-loop gain of 100. Then the new upper cut off frequency is: |
A. | 10 Hz |
B. | 100 Hz |
C. | 10 kHz |
D. | 100 kHz |
Answer» D. 100 kHz | |
45. |
Calculate 0 |
A. | 12 V |
B. | 12 V |
C. | 18 V |
D. | 18 V |
Answer» B. 12 V | |
46. |
Find 0 for the circuit given below: |
A. | |
B. | |
Answer» B. | |
47. |
The circuit is as shown below:
|
A. | 1 |
B. | 1 |
C. | |
D. | 50 |
Answer» B. 1 | |
48. |
If open-loop gain is Aod = 999 in the circuit of the closed-loop gain is: |
A. | 0.999 |
B. | 0.999 |
C. | 1.001 |
D. | 1.001 |
Answer» C. 1.001 | |
49. |
The op-amp of figure has a very poor open loop voltage gain of 45 but is otherwise ideal.
|
A. | 5 |
B. | 20 |
C. | 4 |
D. | 4.5 |
Answer» E. | |
50. |
For the circuit shown below, the output voltage is 0 = 2.5 V in response to input voltage i = 5 V. The finite open-loop differential gain of the op-amp is: |
A. | 5 10 |
B. | 250.5 |
C. | 2 10 |
D. | 501 |
Answer» C. 2 10 | |