The input to a differentiator is –5 V. Its output will be
0 V
A differentiator responds to change, and a steady –5 V does not change — so the output is 0 V, option 2.
\(V_{o}=-RC\dfrac{dV_{i}}{dt}\)
With \(V_{i}=-5\) V constant, \(\dfrac{dV_{i}}{dt}=0\), and the output is zero regardless of the values of R and C or of how large the input happens to be.
The circuit view says the same thing. The op-amp differentiator has a capacitor in series with the input and a resistor in the feedback path. The inverting terminal is a virtual earth, so the capacitor sees the full input voltage across it, and the current it passes is
\(i=C\dfrac{dV_{i}}{dt}\)
A capacitor blocks DC: once the initial transient has died away, a constant applied voltage drives no current. With no current through the feedback resistor there is no drop across it, and the output sits at the virtual-earth potential — zero.
| Input | Output |
|---|---|
| Constant (DC) | 0 V |
| Ramp, slope m | Constant, −RCm |
| Square wave | Sharp spikes at the edges |
| Sine, \(\sin\omega t\) | \(-RC\omega\cos\omega t\) — leads by 90° |
Why the sign of the input is irrelevant. Option 3 tempts by inverting the –5 V to +5 V, which is what an inverting amplifier of unit gain would do. But the differentiator's minus sign applies to the derivative, not to the input level; the DC value never reaches the output at all.
Options 1 and 4 invert the question. A square wave and a sine wave are what a differentiator produces from other inputs — spikes from a square wave, a cosine from a sine — not what it produces from a constant.
A practical footnote. The ideal differentiator's gain rises with frequency as \(\omega RC\), which amplifies high-frequency noise and threatens instability. Real circuits therefore add a small resistor in series with the input capacitor, so that above a chosen corner frequency the stage stops differentiating and behaves as a plain inverting amplifier. That modification does not alter the DC result: the series capacitor still blocks the steady input, and the output is still zero.
Hence, the output is 0 V.
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.
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 :
Assertion (A) : Op-Amp is used for sensor circuit.
Reason (R) : A small signal amplifier amplify weak measured signals.
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 _____________.
