Electronics and Communication questions reward anyone who is comfortable moving between the time domain and the frequency domain. This set covers network theory, analog and digital circuits, signals and systems, control systems, communication systems, and electromagnetics. Numerical solutions keep the units visible at every step, since a dropped factor is the most common reason a correct method still produces a wrong option.
The input impedance of an op-amp in an ideal configuration is assumed to be:
Answer: A
Ideal op-amp assumptions include infinite input impedance (no current drawn at inputs), zero output impedance, and infinite open-loop gain.
Q.22Easy
In a Common Emitter BJT amplifier, if the collector resistance is increased while keeping other parameters constant, how does the voltage gain change?
Answer: A
Voltage gain Av = -gm × Rc, where gm is transconductance and Rc is collector resistance. Increasing Rc directly increases voltage gain.
Q.23Easy
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?
What is the primary advantage of a Darlington pair configuration in BJT amplifiers?
Answer: A
Darlington pairs consist of two transistors in cascade, providing extremely high current gain (β² product) and very high input impedance due to the two base-emitter junctions.
Q.25Easy
A BJT is operated in the saturation region. Which of the following statements is true?
Answer: B
In saturation, both junctions are forward biased: base-emitter junction is forward biased for current conduction, and base-collector junction is also forward biased, allowing maximum collector current.
Q.26Easy
In a two-stage cascaded amplifier, the first stage has gain A₁ = 50 and bandwidth BW₁ = 100 kHz, while the second stage has A₂ = 20 and BW₂ = 200 kHz. What is the overall voltage gain?
Answer: B
In cascaded amplifiers, overall voltage gain = A₁ × A₂ = 50 × 20 = 1000. The bandwidth is determined by the stage with lowest bandwidth (100 kHz).
Q.27Easy
A voltage-controlled voltage source (VCVS) has transconductance gm = 50 mS. If the input voltage is 10 mV, what is the output current?
Answer: A
Output current Id = gm × Vgs = 50 mS × 10 mV = 50 × 10⁻³ × 10 × 10⁻³ = 0.5 × 10⁻³ A = 0.5 mA
Q.28Easy
What is the typical input impedance of a voltage follower (unity gain buffer) using an ideal op-amp?
Answer: B
An ideal op-amp has infinite input impedance. The voltage follower configuration maintains this high input impedance at the non-inverting input.
Q.29Easy
For maximum power transfer from a source with internal resistance Rs to load RL, what should be the condition?
Answer: B
Maximum power transfer theorem states that maximum power is delivered when load impedance equals the complex conjugate of source impedance. For resistive cases, RL = Rs.
Q.30Easy
What is the input offset voltage of a typical precision op-amp (like OP07)?
Answer: C
Precision op-amps like OP07, OPA2134 are designed with input offset voltages < 1 mV. General purpose op-amps (e.g., 741) have offset voltages of 1-5 mV.
Q.31Easy
What is the slew rate of a standard 741 op-amp?
Answer: A
The 741 op-amp has a slew rate of approximately 0.5 V/μs. This limits the maximum output voltage change rate, causing distortion at high frequencies/amplitudes.
Q.32Easy
What is the gain-bandwidth product (GBW) of a typical 741 op-amp?
Answer: A
The 741 op-amp has a gain-bandwidth product of approximately 1 MHz. This means for a gain of 100 V/V, maximum usable bandwidth is ~10 kHz.
Q.33Easy
What is the bandwidth of a first-order low-pass filter with cutoff frequency fc = 1 kHz?
Answer: B
Bandwidth of a low-pass filter is defined by its cutoff frequency (-3dB point). At f = fc, magnitude drops to 0.707 of DC gain. Bandwidth = fc = 1 kHz.
