Integrator MCQ [Free PDF] – Objective Question Answer for Integrator Quiz

1. The circuit in which the output voltage waveform is the integral of the input voltage waveform is called

A. Integrator
B. Differentiator
C. Phase shift oscillator
D. Square wave generator

Answer: A

The integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

 

2. Find the output voltage of the integrator

A. Vo = (1/R×CFt0 Vindt+C
B. Vo = (R/CFt0 Vindt+C
C. Vo = (CF/R)×t0 Vindt+C
D. Vo = (R×CFt0 Vindt+C

Answer: A

The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RCF.

Vo = (1/R×CFt0 Vindt+C

Where C-> Integration constant and CF-> Feedback capacitor.

 

3. Why an integrator cannot be made using a low pass RC circuit?

A. It requires a large value of R and a small value of C
B. It requires a large value of C and a small value of R
C. It requires a large value of R and C
D. It requires a small value of R and C

Answer: C

A simple low pass RC circuit can work as an integrator when the time constant is very large, which requires the large value of R and C. Due to practical limitations, the R and C cannot be made infinitely large.

 

4. How a perfect integration is achieved in op-amp?

A. Infinite gain
B. Low input impedance
C. Low output impedance
D. High CMRR

Answer: A

In an op-amp integrator, the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for the ideal op-amp. So, the effective time constant of the op-amp integrator becomes very large which results in perfect integration.

 

5. The op-amp operating in open-loop results in the output of the amplifier saturating at a voltage

A. Close to op-amp positive power supply
B. Close to op-amp negative power supply
C. Close to op-amp positive or negative power supply
D. None of the mentioned

Answer: C

In practice, the output of the op-amp never becomes infinite rather the output of the op-amp saturates at a voltage close to the op-amp’s positive or negative power supply depending on the polarity of the input dc signal.

 

6. The frequency at which gain is 0db for the integrator is

A. f=1/(2πRFCF)
B. f=1/(2πR1CF)
C. f=1/(2πR1R1)
D. f=(1/2π)×(RF/R1)

Answer: B

The frequency at which the gain of the integrator becomes zero is f=1/(2πR1CF).

 

7. Why practical integrator is called a lossy integrator?

A. Dissipation power
B. Provide stabilization
C. Changes input
D. None of the mentioned

Answer: D

To avoid saturation problems, the feedback capacitor is shunted by a feedback resistance(RF). The parallel combination of RF and CF behaves like a practical capacitor that dissipates power. For this reason, the practical integrator is called a lossy integrator.

 

8. Determine the lower frequency limit of integration for the circuit given below.
Determine the lower frequency limit of integration for the circuit given below.
A. 43.43kHz
B. 4.82kHz
C. 429.9kHz
D. 4.6MHz

Answer: B

The lower frequency limit of integration,

f= 1/(2πRFCF)

= 1/(2π×1kΩ×33nF)

= 4.82kHz.

 

9. Find the range of frequency between which the circuit act as an integrator?

A. [1/(2πRFCF)]– (2πR1CF)
B. (2πRFCF) – [1/(2πR1CF)].
C. [1/(2πRFCF)]- [1/(2πR1CF)].
D. None of the mentioned

Answer: C

The range of frequency between which the circuit act as an integrator as

[1/(2πRFCF)]- [1/(2πR1CF)].

 

10. What will be the output voltage waveform for the circuit, R1×CF=1s and input is a step voltage. Assume that the op-amp is initially nulled.

A. Triangular function
B. Unit step function
C. Ramp function
D. Square function

Answer: C

Input voltage Vin = 1.2v for 0≤t≤0.4ms. The output voltage at t=0.4ms is

Vo = (1/R×CFt0 Vindt+C =-(1/1) × 0.401.2 dt

=> Vo =-[0.101.2 dt + 0.20.11.2 dt + 0.30.21.2 dt + 0.40.31.2 dt ] = –

(1.2+1.2+1.2+1.2) = -4.8v

Therefore, the output voltage waveform is a ramp function.

Scroll to Top