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| By Suddm.kgp
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Suddm.kgp
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Quizzes Created: 2 | Total Attempts: 625
Questions: 20 | Attempts: 394

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• 1.

### When a differential amplifier is operated single-ended,…….

• A.

The output is grounded

• B.

One input is grounded and signal is applied to the other

• C.

Both inputs are connected together

• D.

The output is not inverted

B. One input is grounded and signal is applied to the other
Explanation
When a differential amplifier is operated single-ended, one input is grounded and the signal is applied to the other. This means that one input of the differential amplifier is connected to the reference ground, while the signal is applied to the other input. This configuration allows for the amplification of the voltage difference between the two inputs, which is useful in many applications such as amplifying small signals or rejecting common-mode noise.

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• 2.

### In differential-mode, …………….

• A.

Opposite polarity signals are applied to the inputs

• B.

The gain is one

• C.

The outputs are of different amplitudes

• D.

Only one supply voltage is used

A. Opposite polarity signals are applied to the inputs
Explanation
In differential-mode, opposite polarity signals are applied to the inputs. This means that one input receives a positive signal while the other receives a negative signal. This configuration allows for the cancellation of common-mode noise, as any noise that is present in both signals will be subtracted out. This is useful in applications where noise rejection is important, such as in audio or communication systems.

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• 3.

### In the common mode,

• A.

Both inputs are grounded

• B.

The outputs are connected together

• C.

An identical signal appears on both the inputs

• D.

The output signal are in-phase

C. An identical signal appears on both the inputs
Explanation
In the common mode, an identical signal appears on both the inputs. This means that both inputs receive the same signal with the same amplitude and phase. This can occur when there is a common source or noise that affects both inputs equally. In this scenario, the outputs will also be connected together and will have the same signal, resulting in the output signals being in-phase.

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• 4.

• A.

80dB

• B.

40dB

• C.

20dB

• D.

100dB

A. 80dB
• 5.

### The slew rate of an op-amp whose output increases 8V in 12μs is

• A.

90 V/μs

• B.

0.67 V/μs

• C.

1.5 V/μs

• D.

None of these

B. 0.67 V/μs
Explanation
The slew rate of an op-amp is the maximum rate at which its output voltage can change. In this question, the output voltage increases by 8V in 12μs. To find the slew rate, we divide the change in voltage by the time it takes, giving us a rate of approximately 0.67 V/μs. Therefore, the correct answer is 0.67 V/μs.

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• 6.

### For an Op-amp with negative feedback, the output is

• A.

Equal to the input

• B.

Increased

• C.

Fed back to the inverting input

• D.

Fed back to the non-inverting input

C. Fed back to the inverting input
Explanation
In an operational amplifier (op-amp) with negative feedback, the output is fed back to the inverting input. This feedback mechanism helps to stabilize the amplifier's gain and reduce distortion. By feeding the output back to the inverting input, the op-amp adjusts its output to minimize the difference between the inverting and non-inverting inputs. This causes the output to be proportional to the input signal, resulting in a high gain and accurate amplification.

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• 7.

### A certain inverting amplifier has a closed-loop voltage gain of 25. The Op-amp has an open-loop voltage gain of 100,000. If an Op-amp with an open-loop voltage gain of 200,000 is substituted in the arrangement, the closed-loop gain

• A.

Doubles

• B.

Drops to 12.5

• C.

Remains at 25

• D.

Increases slightly

C. Remains at 25
Explanation
The closed-loop gain of an inverting amplifier is determined by the ratio of the feedback resistor to the input resistor. It is independent of the open-loop gain of the operational amplifier. Therefore, substituting an op-amp with a higher open-loop gain will not affect the closed-loop gain of the amplifier. The closed-loop gain will remain at 25.

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• 8.

### The Op-amp can amplify

• A.

A.c. signals only

• B.

D.c. signals only

• C.

Both a.c. and d.c. signals

• D.

Neither d.c. nor a.c. signals

C. Both a.c. and d.c. signals
Explanation
An operational amplifier (Op-amp) is a device that can amplify both alternating current (a.c.) and direct current (d.c.) signals. It is designed to have a high gain and can amplify signals with frequencies ranging from zero to very high frequencies. This makes it suitable for amplifying both a.c. signals, which have varying amplitudes and frequencies, and d.c. signals, which have constant amplitudes. Therefore, the Op-amp is capable of amplifying both types of signals, making the correct answer "both a.c. and d.c. signals".

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• 9.

### The current flowing through emitter resistance of a differential amplifier is …….

• A.

Half of either collector current

• B.

Equal to either collector current

• C.

Two times either collector current

• D.

Equal to the difference in base currents

C. Two times either collector current
Explanation
The current flowing through the emitter resistance of a differential amplifier is two times either collector current. This is because in a differential amplifier, the emitter current is divided equally between the two transistors. Therefore, each collector current is equal to half of the emitter current. Since the current flowing through the emitter resistance is the sum of the collector currents, it will be two times either collector current.

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• 10.

### An astable multivibrator is also known as a

• A.

One-shot multivibrator

• B.

Free-running multivibrator…

• C.

Bistable multivibrator

• D.

Monostable multivibrator

B. Free-running multivibrator…
Explanation
An astable multivibrator is a type of electronic circuit that continuously oscillates between two states without any external triggering. It is commonly known as a free-running multivibrator because it operates on its own without any external control or input. The circuit uses a combination of capacitors, resistors, and transistors to create a continuous oscillation of the output waveform. This makes it useful in applications such as timing circuits, clock generators, and frequency dividers.

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• 11.

### What starts a free-running multivibrator?

• A.

An input signal

• B.

A trigger

• C.

An external circuit

• D.

Nothing

D. Nothing
Explanation
A free-running multivibrator is a circuit that generates a continuous output waveform without the need for any external input or trigger. It is also known as an astable multivibrator. Unlike other types of multivibrators, it does not require any external signal or circuit to initiate its operation. It is designed to oscillate continuously, generating a square wave or pulse waveform without any external stimulus. Therefore, the correct answer is "Nothing".

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• 12.

### What is the cutoff frequency of the low pass filter in the following circuit

• A.

4.8 kHz

• B.

3.8 kHz

• C.

2.8 kHz

• D.

1.8 kHz

A. 4.8 kHz
Explanation
The cutoff frequency of a low pass filter determines the frequency at which the filter starts attenuating the input signal. In this case, the correct answer is 4.8 kHz, indicating that the low pass filter in the circuit begins to attenuate frequencies above 4.8 kHz.

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• 13.

### What is the frequency of this 555 astable multivibrator?

• A.

241 Hz

• B.

178 Hz

• C.

241 kHz

• D.

178 kHz

A. 241 Hz
Explanation
The frequency of a 555 astable multivibrator is determined by the values of the resistors and capacitors connected to it. In this case, the frequency is 241 Hz, indicating that the combination of resistors and capacitors used in the circuit produces a square wave output with a frequency of 241 cycles per second.

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• 14.

### If the input to a comparator is a sine wave, the output is a

• A.

Ramp voltage

• B.

Sine wave

• C.

Rectangular wave

• D.

Sawtooth wave

C. Rectangular wave
Explanation
A comparator is a device that compares two input voltages and produces an output based on the comparison. In this case, if the input to the comparator is a sine wave, the comparator will compare the varying amplitude of the sine wave with a reference voltage. Whenever the amplitude of the sine wave exceeds the reference voltage, the output of the comparator will switch to a high state, and when the amplitude is below the reference voltage, the output will switch to a low state. This behavior of switching between high and low states produces a rectangular wave as the output.

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• 15.

### The Schmitt trigger is a two-state device used for

• A.

Pulse shaping

• B.

Peak detection

• C.

Input noise rejection

• D.

Filtering

A. Pulse shaping
Explanation
The Schmitt trigger is a two-state device that is commonly used for pulse shaping. It is designed to convert a noisy or distorted input signal into a clean and well-defined square wave output signal. By using positive feedback, the Schmitt trigger can effectively eliminate small fluctuations and noise in the input signal, ensuring that only significant changes in voltage trigger a state transition. This makes it ideal for applications such as debouncing mechanical switches or generating clean digital signals from analog inputs. Therefore, pulse shaping is the correct answer.

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• 16.

### A circuit that uses an amplifier with passive filter elements is called a/an:

• A.

Relaxation oscillator

• B.

Signal generator

• C.

Active filter

• D.

Differentiator

C. Active filter
Explanation
An active filter is a circuit that uses an amplifier along with passive filter elements such as resistors, capacitors, and inductors to filter out specific frequencies from a signal. Unlike passive filters which only use passive components, an active filter can provide gain and can be designed to have a desired frequency response. Therefore, an active filter is the correct answer for a circuit that uses an amplifier with passive filter elements.

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• 17.

### A low pass filter is one that passes all the frequencies from

• A.

A fixed frequency to ∞

• B.

0 to fixed higher frequency

• C.

A fixed lower frequency to fixed higher frequency

• D.

All frequencies

B. 0 to fixed higher frequency
Explanation
A low pass filter is a type of filter that allows frequencies below a certain cutoff frequency to pass through while attenuating frequencies above the cutoff. In this case, the correct answer is "0 to fixed higher frequency," indicating that the filter allows all frequencies from 0 Hz up to a fixed higher frequency to pass through.

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• 18.

### A high pass filter is one that passes all the frequencies from

• A.

A fixed frequency to ∞

• B.

0 to a fixed higher frequency

• C.

A fixed lower frequency to fixed higher frequency

• D.

All frequencies

A. A fixed frequency to ∞
Explanation
A high pass filter is designed to allow frequencies above a certain fixed frequency to pass through while attenuating frequencies below that frequency. In other words, it passes all frequencies from a fixed frequency to infinity, meaning that any frequency higher than the fixed frequency will not be attenuated or filtered out.

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• 19.

### A passive filter makes use of

• A.

RC networks

• B.

RLC networks

• C.

LC networks

• D.

All of the above

D. All of the above
Explanation
A passive filter makes use of RC networks, RLC networks, and LC networks. RC networks consist of resistors and capacitors, RLC networks consist of resistors, inductors, and capacitors, and LC networks consist of inductors and capacitors. Each type of network has its own characteristics and is used in different applications to filter out specific frequencies or signals. Therefore, a passive filter can utilize any combination of these networks depending on the desired filtering requirements.

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• 20.

### Calculate the cutoff frequency of a first-order low-pass filter for R1 = 2.5 kΩ and C1 = 0.05 µF

• A.

1.273 kHz

• B.

12.73kHz

• C.

127.3kHz

• D.

1237 kHz.

A. 1.273 kHz
Explanation
The cutoff frequency of a first-order low-pass filter can be calculated using the formula f = 1 / (2πRC), where f is the cutoff frequency, R is the resistance, and C is the capacitance. Plugging in the given values of R1 = 2.5 kΩ and C1 = 0.05 µF into the formula, we get f = 1 / (2π * 2.5 * 10^3 * 0.05 * 10^-6) = 1.273 kHz. Therefore, the correct answer is 1.273 kHz.

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• Current Version
• Mar 03, 2023
Quiz Edited by
ProProfs Editorial Team
• Apr 25, 2019
Quiz Created by
Suddm.kgp

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