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Electronics (ECE)

Analog/digital electronics, communication

135 Q 4 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 51–60 of 135
Topics in Electronics (ECE)
Q.51 Easy Analog Circuits
A voltage follower (Common Collector) amplifier has a voltage gain of:
A Approximately 1
B β (current gain)
C Greater than 1
D Approximately -1
Correct Answer:  A. Approximately 1
EXPLANATION

The voltage gain of a voltage follower is Av ≈ 1 (or slightly less due to emitter resistance effects). It is used for impedance matching.

Test
Q.52 Easy Analog Circuits
In a differential amplifier, the Common Mode Rejection Ratio (CMRR) is 80 dB. What is the CMRR in linear form?
A 10,000
B 100,000
C 1,000
D 80
Correct Answer:  A. 10,000
EXPLANATION

CMRR (dB) = 20 log(CMRR linear). Therefore, 80 = 20 log(CMRR), giving CMRR = 10^4 = 10,000.

Test
Q.53 Easy Analog Circuits
The input impedance of a Common Emitter amplifier is approximately equal to:
A β × re
B 1/(gm)
C RC || RL
D RE only
Correct Answer:  A. β × re
EXPLANATION

The input impedance of a CE amplifier is Zin ≈ β × re, where β is current gain and re is emitter resistance (≈ VT/IE).

Test
Q.54 Easy Analog Circuits
A BJT amplifier operates in the saturation region. What is the approximate output impedance?
A Very high (MΩ range)
B Very low (few ohms)
C Moderate (kΩ range)
D Undefined in saturation
Correct Answer:  B. Very low (few ohms)
EXPLANATION

In saturation, the transistor acts almost like a closed switch with minimal voltage drop across it, resulting in very low output impedance.

Test
Q.55 Easy Analog Circuits
In a Common Source FET amplifier, the gain is primarily dependent on which parameter?
A Transconductance (gm) and load resistance (RD)
B Only the gate-source voltage
C Drain-source resistance only
D Channel length modulation factor only
Correct Answer:  A. Transconductance (gm) and load resistance (RD)
EXPLANATION

The voltage gain of a CS amplifier is Av = -gm × RD, where gm is transconductance and RD is the drain resistance. This is the fundamental relationship.

Test
Q.56 Easy Analog Circuits
A power amplifier delivers 100 W to a 8 Ω load. What is the RMS output voltage? (Assuming pure resistive load)
A 28.3 V
B 40 V
C 20 V
D 14.1 V
Correct Answer:  A. 28.3 V
EXPLANATION

P = V²/R → V = √(P×R) = √(100×8) = √800 ≈ 28.3 V RMS.

Test
Q.57 Easy Analog Circuits
A transistor amplifier has input signal of 10 mV and output signal of 5 V. What is the voltage gain in dB?
A 54 dB
B 40 dB
C 68 dB
D 500 dB
Correct Answer:  A. 54 dB
EXPLANATION

Gain = Vout/Vin = 5V/10mV = 500. Gain (dB) = 20 log₁₀(500) ≈ 54 dB.

Test
Q.58 Easy Analog Circuits
For an ideal op-amp, the input offset voltage is:
A Maximum possible
B Zero
C Dependent on supply voltage
D 1 mV
Correct Answer:  B. Zero
EXPLANATION

An ideal op-amp has zero input offset voltage, zero input bias current, infinite gain, infinite input impedance, and zero output impedance.

Test
Q.59 Easy Analog Circuits
Which type of feedback is applied in a non-inverting amplifier configuration?
A Positive feedback
B Negative feedback
C No feedback
D Regenerative feedback
Correct Answer:  B. Negative feedback
EXPLANATION

Non-inverting amplifier uses negative feedback from output to inverting input, which reduces gain but improves linearity, bandwidth, and input impedance.

Test
Q.60 Easy Analog Circuits
In a Hartley oscillator, the frequency of oscillation is determined by which circuit element combination?
A Inductor and capacitor only
B Two inductors and one capacitor (LC tank)
C Resistance and inductance
D Two capacitors and resistance
Correct Answer:  B. Two inductors and one capacitor (LC tank)
EXPLANATION

Hartley oscillator uses two inductors and one capacitor in the tank circuit. Frequency f = 1/(2π√L_eq×C) where L_eq is equivalent inductance.

Test
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