# Analog Electronics - 6

## 10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Analog Electronics - 6

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Attempt Analog Electronics - 6 | 10 questions in 30 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
QUESTION: 1

### The current flowing through the 3 Ω resistor R1 is _____ A. Solution: Apply nodal at A: Magnitude wise: 2 A

*Answer can only contain numeric values
QUESTION: 2

### An op-amp has a voltage gain of 100 dB at dc and a unity gain frequency of 5 MHz. The lower 3-dB cut off frequency is _______ Hz.

Solution:

100 dB voltage gain

20 log (Gain) = 100

Gain = 105

GBW = 5 MHz

Low frequency 3-dB = = 50 Hz.

*Answer can only contain numeric values
QUESTION: 3

### An op-amp with slew rate 1 V/μ is used in the circuit below input step voltage Vi = V sin (105t) is given, the maximum value of V such that no raise time distortion occurs at the output is _____V. Solution:

The output Vo = -3V sin (ω0t) For no distortion to occur

3Vω0 ≤ SR *Answer can only contain numeric values
QUESTION: 4

The circuit shows an analog voltmeter of very high input impedance that uses an inexpensive moving coil-meter. The voltmeter measures the voltage ‘V’ applied between the op-amp’s positive-input terminal and ground. Assuming that the moving coil produces full-scale deflection when the current passing through it is 100 μA, the value of R (in kΩ) to obtain full-scale reading at +10 V is Solution: VA = V (virtual ground) QUESTION: 5

Circuit shows an op-amp circuit used for generating the square waveform. If the output frequency required in 1 kHz. then the possible values of R1­ and R2 can be respectively

Solution:

Frequency of oscillation If R1 = 2k

R2 = 0.859 × 2

= 1.7 k

Option 1 is correct.

*Answer can only contain numeric values
QUESTION: 6

The circuit shows the series voltage regulator. If reference voltage Vref is 2V and R1 = 3K, R2 = 1K, β = 99 for both transistors. Then the value of current I0 in micro-amperes is _____ μA.

Solution:

V- = V+ = 2V  = 2 mA

I1 = (1 + β) (1 + β) I0

2 mA = 104 I0 I = 0.2 uA

QUESTION: 7

The circuit shows an – op-amp in Schmidt Trigger configuration. If the hysteresis width is 3V. Then the value of R1 and R2 is Solution:   *Answer can only contain numeric values
QUESTION: 8

For the amplifier circuit shown, the op-amp can deliver a maximum current of 100 mA and is powered by ±15 V supply. If the input to the amplifier is a triangular waveform as shown. The peak value of the output waveform is ______. Solution:

The circuit is non-inverting.

Amplifier: = 11V

However the current flowing through load when V0 = 11 V is Which is more than the current op-amp can supply.

Thus peak value is limited by the maximum current:

50 Ω ≪ 110 kΩ

All the current flows through 50 Ω resistor.

V0 = (100 mA) (50 Ω)

= 5 V

QUESTION: 9

An op-amp circuit is shown in the figure. If Vin is an AC source of frequency ω. Consider the following statements:
a) For ω Circuit is an integrator
(b) For ω circuit is an Amplifier

Solution:   Circuit acts like an Integrator Circuit acts an Inverting Amplifier

The given conditions in question is opposite Hence None of a and b is correct

*Answer can only contain numeric values
QUESTION: 10

The average power delivered by the 3 V source is ________ mW. Solution: V- = V+ = 3 V [Virtual ground]

Current through 1 k resistor

I = 1 mA

Voltage across 1k resistor from voltage division

Voltage across  Vo = 15 V

KVL at V+ node:

3 – I'(8) = Vo

3 – I'(8) = 15

3 – 15 = I'(8)  Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code