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

2. Find the output voltage of the integrator

A. V_{o} = (1/R×C_{F})×^{t}∫_{0} V_{in}dt+C

B. V_{o} = (R/C_{F})×^{t}∫_{0} V_{in}dt+C

C. V_{o} = (C_{F}/R)×^{t}∫_{0} V_{in}dt+C

D. V_{o} = (R×C_{F})×^{t}∫_{0} V_{in}dt+C

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

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

A. Infinite gain

B. Low input impedance

C. Low output impedance

D. High CMRR

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

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

A. f=1/(2πR_{F}C_{F})

B. f=1/(2πR_{1}C_{F})

C. f=1/(2πR_{1}R_{1})

D. f=(1/2π)×(R_{F}/R_{1})

7. Why practical integrator is called a lossy integrator?

A. Dissipation power

B. Provide stabilization

C. Changes input

D. None of the mentioned

8. Determine the lower frequency limit of integration for the circuit given below.

A. 43.43kHz

B. 4.82kHz

C. 429.9kHz

D. 4.6MHz

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

A. [1/(2πR_{F}C_{F})]– (2πR_{1}C_{F})

B. (2πR_{F}C_{F}) – [1/(2πR_{1}C_{F})].

C. [1/(2πR_{F}C_{F})]- [1/(2πR_{1}C_{F})].

D. None of the mentioned

10. What will be the output voltage waveform for the circuit, R_{1}×C_{F}=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