
An op amp has an offset voltage of 1mV and is ideal in all other respects. If this op amp is used in the circuit shown in figure the output voltage would be:

(A) 1 mV
(B) 1 V
(C) \[0.1{\text{V}}\]
(D) 0 V.
Answer
135k+ views
Hint: In such a circuit, the offset voltage (also known as the input offset voltage) is the input voltage to the op amp. The ratio of the output voltage to the input voltage is equal to the ratio of the resistance across the op amp and the input resistance of the op amp
Formula used: In this solution we will be using the following formulae;
\[{V_{out}} = \dfrac{{{R_F}}}{{{R_{in}}}}{V_{off}}\] where \[{V_{out}}\] is the output voltage of the op amp, \[{R_F}\] is the resistance connected across the op amp, \[{R_{in}}\] is the input resistance usually connected to the negative part terminal of the op amp, and \[{V_{off}}\] is the offset voltage.
Complete Step-by-Step solution:
To solve the above question, as told, we note that the op amp is ideal. This means that for such a circuit the offset voltage would be amplified. This offset voltage is a parameter of an amplifier which gives information about the required input voltage for the operation of the amplifier.
Generally, the ratio of the output voltage to the input voltage is equal to the ratio of the resistance across the op amp and the input resistance of the op amp
\[{V_{out}} = \dfrac{{{R_F}}}{{{R_{in}}}}{V_{off}}\] where \[{V_{out}}\] is the output voltage of the op amp, \[{R_F}\] is the resistance connected across the op amp, \[{R_{in}}\] is the input resistance usually connected to the negative part terminal of the op amp, and \[{V_{off}}\] is the offset voltage.
By inserting known values, we have
\[{V_{out}} = \dfrac{{1000k\Omega }}{{1k\Omega }}0.001{\text{V}} = 1{\text{V}}\]
Hence, the correct option is B.
Note: As seen from the diagram, there is no input signal in the system as both the negative and positive are grounded. Hence, why the offset voltage is used as the input. Also, this is ideal, an non-ideal op amp may give a zero volt at output even at a non zero offset voltage.
Formula used: In this solution we will be using the following formulae;
\[{V_{out}} = \dfrac{{{R_F}}}{{{R_{in}}}}{V_{off}}\] where \[{V_{out}}\] is the output voltage of the op amp, \[{R_F}\] is the resistance connected across the op amp, \[{R_{in}}\] is the input resistance usually connected to the negative part terminal of the op amp, and \[{V_{off}}\] is the offset voltage.
Complete Step-by-Step solution:
To solve the above question, as told, we note that the op amp is ideal. This means that for such a circuit the offset voltage would be amplified. This offset voltage is a parameter of an amplifier which gives information about the required input voltage for the operation of the amplifier.
Generally, the ratio of the output voltage to the input voltage is equal to the ratio of the resistance across the op amp and the input resistance of the op amp
\[{V_{out}} = \dfrac{{{R_F}}}{{{R_{in}}}}{V_{off}}\] where \[{V_{out}}\] is the output voltage of the op amp, \[{R_F}\] is the resistance connected across the op amp, \[{R_{in}}\] is the input resistance usually connected to the negative part terminal of the op amp, and \[{V_{off}}\] is the offset voltage.
By inserting known values, we have
\[{V_{out}} = \dfrac{{1000k\Omega }}{{1k\Omega }}0.001{\text{V}} = 1{\text{V}}\]
Hence, the correct option is B.
Note: As seen from the diagram, there is no input signal in the system as both the negative and positive are grounded. Hence, why the offset voltage is used as the input. Also, this is ideal, an non-ideal op amp may give a zero volt at output even at a non zero offset voltage.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage
