121. Which circuit can be used to take the square root of a signal?
A. Divider circuit
B. Multiplier circuit
C. Squarer circuit
D. None of the mentioned
122. Find the output voltage of the squarer circuit?
A. Vo = √( Vref/|Vin|)
B. Vo = √( Vref×|Vin|)
C. Vo = √( Vref2×|Vin|)
D. Vo = √( Vref×|Vin|2)
123. Find the current, IL flowing in the circuit given below
A. IL = 1.777mA
B. IL = 1.048mA
C. IL = 1.871mA
D. None of the mentioned
124. A square root circuit built from the multiplier is given an input voltage of 11.5v. Find its corresponding output voltage?
A. 11v
B. 15v
C. 13v
D. Cannot be determined
125. 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
126. Find the output voltage of the integrator
A. Vo = (1/R×CF)×t∫0 Vindt+C
B. Vo = (R/CF)×t∫0 Vindt+C
C. Vo = (CF/R)×t∫0 Vindt+C
D. Vo = (R×CF)×t∫0 Vindt+C
127. 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
128. How a perfect integration is achieved in op-amp?
A. Infinite gain
B. Low input impedance
C. Low output impedance
D. High CMRR
129. 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
130. 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)
131. Why practical integrator is called a lossy integrator?
A. Dissipation power
B. Provide stabilization
C. Changes input
D. None of the mentioned
132. Determine the lower frequency limit of integration for the circuit given below.
A. 43.43kHz
B. 4.82kHz
C. 429.9kHz
D. 4.6MHz
133. 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
134. 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
135. Find R1 and RF in the lossy integrator so that the peak gain and the gain down from its peak is 40db to 6db. Assume ω=20,000 rad/s and capacitance = 0.47µF.
A. R1 = 10.6Ω, RF = 106Ω
B. R1 = 21.2Ω, RF = 212.6Ω
C. R1 = 42.4Ω, RF = 424Ω
D. R1 = 29.8Ω, RF = 298Ω
136. Why a resistor is shunted across the feedback capacitor in the practical integrator?
A. To reduce the operating frequency
B. To enhance low-frequency gain
C. To enhance error voltage
D. To reduce error voltage
137. Find the application in which integrator is used?
A. All of the mentioned
B. Analog Computers
C. FM Detectors
D. AM detectors
138. At what condition the input signal of the integrator is integrated properly
A. T = RFCF
B. T ≤ RFCF
C. T ≥ RFCF
D. T ≠ RFCF
139. What happens if the input frequency is kept lower than the frequency at which the gain is zero?
A. Circuit acts like a perfect integrator
B. Circuit acts like an inverting amplifier
C. Circuit acts like a voltage follower
D. Circuit acts like a differentiator
140. Match the correct frequency range for integration. (Where f –> Input frequency and fa –> Lower frequency limit of integration)
1. f a | i. Results in 50% accuracy in integration |
2.f = fa | ii. Results in 99% accuracy in integration |
3.f = 10fa | iii. No integration results |
A. 1-iii, 2-i, 3-ii
B. 1-i, 2-ii, 3-iii
C. 1-ii, 2-iii, 1-i
D. 1-iii, 2-ii, 3-i