Assertion (A) : Op-Amp is used for sensor circuit. Reason (R) : A small signal amplifier amplify weak measured signals.
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both are true, and the reason states exactly why the assertion holds — option 1.
How small sensor outputs actually are.
| Sensor | Output |
|---|---|
| Thermocouple | ~40 µV per °C |
| Strain gauge bridge | A few mV at full scale |
| ECG electrode | 1 – 2 mV |
| Photodiode | Nanoamps to microamps |
An ADC expects volts. Between the two sits an amplifier, and the op-amp is what fills the gap — usually as an instrumentation amplifier built from three of them.
Amplification is necessary but not sufficient, and the properties that make the op-amp the right choice all matter here:
Differential input with high CMRR. A bridge or a pair of electrodes delivers a small difference riding on a large common-mode voltage — mains hum, supply offset, the bridge's own excitation. A CMRR above 100 dB rejects what is common and keeps only the difference. A single-ended amplifier could not.
High input impedance. A sensor with a source impedance of megohms — a pH electrode, a piezoelectric transducer — would be loaded and its output attenuated by anything less. The op-amp draws picoamps.
Low offset and drift. Amplifying 40 µV by a thousand also amplifies the amplifier's own input offset voltage by a thousand. A chopper-stabilised device with 1 µV offset and nanovolt-per-degree drift is what makes microvolt measurement possible at all.
Direct coupling. Most sensor signals are DC or very slow, so the amplifier must respond down to zero frequency — which an op-amp does and a capacitively coupled amplifier does not.
The linkage is direct. The reason says weak signals need amplifying; the assertion says op-amps are used in sensor circuits; and it is precisely because sensor signals are weak that an op-amp is placed there. R explains A.
Beyond amplification, the same device also filters, linearises, converts current to voltage, and drives the ADC — but the reason names the primary function correctly.
Hence, both (A) and (R) are true and (R) is the correct explanation of (A).
Similar Questions
Assertion (A) : An Op-Amp is a direct coupled high gain amplifier.
Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
The input to a differentiator is –5 V. Its output will be
Match the following :
List – I List – II a. h-parameters i. O/P voltage varies as the slope of i/p voltage b. differentiator ii. Noise division c. half-wave rectifier iii. Function of a Q point d. integrator iv. series diode clipper Codes :
Consider the following circuit, the switch S1 allows the output to switch between two ranges of amplitudes from 0-0.1 V and 0-1 V. Arrange these values of R1, R2 and R3 in increasing order.
(A) Value of R1
(B) Value of R2
(C) Value of R3Choose the most appropriate answer from the options given below :
The given operational amplifier circuit corresponds to which electronic circuit application ?
Statements in connection to Op-Amp applications are :
A. If we use a square wave generator followed by integrator circuit we get a triangular wave at the output
B. The logarithmic amplifier called a log-amplifier or a logger, is basically a current to voltage converter.
C.
is a first order high pass filter with voltage followerD. If we use a square wave generator followed by a clipping circuit then we get a saw-tooth wave generator.
Choose the correct answer from the options given below:
Match List I with List II
LIST I LIST II A. Single polarity positive voltage clipper circuit I. 
B. Negative clamping circuit II. 
C. Differentiator circuit III. 
D. Logarithmic Amplifier IV. 
Choose the correct answer from the options given below:
The correct flow in the basic block diagram of op-Amp is
A. Complementary symmetry push pull amplifier block.
B. Dual input unbalanced output differential amplifier block
C. Emitter follower with constant current source block
D. Dual-Input balanced output differential amplifier block
Choose the correct answer from the options given below :
Match the following :
List - I List - II a) 
i) 
b) 
ii) 
c) 
iii) 
d) 
iv) 
Codes :
\(\dfrac{d^{2}y}{dt^{2}}+y=0\) can be solved using an Operational Amplifier. Which one is preferred ?
Important Questions from Op-Amp and Its Applications
In the circuit shown below, $R$ is $1\text{ k}\Omega$ and $R_f$ is $10\text{ k}\Omega$. If $V_{\text{in}}$ is $100\text{ mV}$ and the op-amps have supply voltages of $\pm 15\text{ V}$, then $V_{\text{out}}$ is _______ V.
For IC 741 circuit shown below, the maximum input frequency above which the output will be distorted is ____________.
Assume that an op-amp 741 connected as a unit gain inverting amplifier is applied with an input change of 10 V. Determine the time taken for the output to change by 10 V. [Slew rate for op-amp 741 = 0.5 V/µs]
What is the value of output voltage $V_o$ in the given circuit? Assume op-amp to be ideal and MOSFET operating in saturation region.
In an ideal op-amp, the common mode gain is _____________.
