Home Subjects Electronics (ECE) Analog Circuits

Electronics (ECE)
Analog Circuits

Analog/digital electronics, communication

33 Q 4 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 11–20 of 33
Topics in Electronics (ECE)
Q.11 Easy Analog Circuits
A non-inverting amplifier is constructed with an op-amp having open-loop gain A₀ = 100,000. If Rf = 90 kΩ and Rin = 10 kΩ, what is the closed-loop voltage gain?
A 10
B 9
C 11
D 100
Correct Answer:  A. 10
EXPLANATION

For non-inverting amplifier: Acl = 1 + (Rf/Rin) = 1 + (90/10) = 1 + 9 = 10

Test
Q.12 Easy Analog Circuits
In a Common Emitter BJT amplifier, if the collector resistance is increased while keeping other parameters constant, how does the voltage gain change?
A Voltage gain increases
B Voltage gain decreases
C Voltage gain remains constant
D Voltage gain becomes zero
Correct Answer:  A. Voltage gain increases
EXPLANATION

Voltage gain Av = -gm × Rc, where gm is transconductance and Rc is collector resistance. Increasing Rc directly increases voltage gain.

Test
Q.13 Easy Analog Circuits
The input impedance of an op-amp in an ideal configuration is assumed to be:
A Infinite
B Zero
C Equal to the feedback resistance
D 1 MΩ
Correct Answer:  A. Infinite
EXPLANATION

Ideal op-amp assumptions include infinite input impedance (no current drawn at inputs), zero output impedance, and infinite open-loop gain.

Test
Q.14 Easy Analog Circuits
The gain-bandwidth product (GBW) of a typical general-purpose op-amp (like LM741) is approximately:
A 1 MHz
B 10 MHz
C 100 MHz
D 1 GHz
Correct Answer:  A. 1 MHz
EXPLANATION

LM741 has a GBW product of approximately 1 MHz, which is why it's limited to low-frequency applications. Modern op-amps have higher GBW.

Test
Q.15 Easy Analog Circuits
An op-amp integrator circuit has R = 10 kΩ and C = 100 nF. What is the time constant?
A 1 ms
B 10 ms
C 100 μs
D 1 μs
Correct Answer:  A. 1 ms
EXPLANATION

Time constant τ = RC = 10×10³ × 100×10⁻⁹ = 10×10⁻⁴ = 1×10⁻³ s = 1 ms.

Test
Q.16 Easy Analog Circuits
In a BJT amplifier biased in the active region, what is the relationship between collector current and base current?
A IC = β × IB
B IC = IB
C IC = VCE/RC
D IC is independent of IB
Correct Answer:  A. IC = β × IB
EXPLANATION

In the active region, IC = β × IB where β is the DC current gain (typically 50-300 for silicon BJTs).

Test
Q.17 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.18 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.19 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.20 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
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