How does the varying resistor affect the voltage at node 2?

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Discussion Overview

The discussion revolves around the effects of a varying resistor in an op-amp circuit, specifically focusing on how it influences the voltage at node 2. Participants explore concepts related to nodal analysis, ideal op-amp assumptions, and the role of the varying resistor as a potential divider.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the circuit diagram and the connections of the 4.7k ohm resistor and the 100 ohm varying resistor, questioning how to perform nodal analysis at node 3.
  • Another participant questions whether the op-amp is ideal, suggesting that it may not be for laboratory purposes, although a later reply confirms it is assumed to be ideal.
  • A participant explains that the varying resistor acts as a potentiometer or potential divider, indicating that the voltage at node 2 depends on the position of the dial, with specific voltage outcomes based on its position relative to nodes 1 and 3.
  • The same participant suggests using the potential divider formula to find the voltage at node 2, depending on the resistances involved.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the connections in the circuit and the implications of the varying resistor. There is a mix of assumptions about the ideality of the op-amp, with some participants agreeing on its ideal status while others raise concerns about practical considerations.

Contextual Notes

Participants do not fully resolve the nodal analysis question, and there are missing details regarding the circuit parameters and the specific values of resistances involved. The discussion also reflects differing interpretations of the varying resistor's role in the circuit.

mathrocks
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I have to solve an op amp circuit for a lab I'm doing but the circuit they have given me looks confusing (it's on page 46, http://filebox.vt.edu/users/oshekari/Manual_Student.pdf). Why does the 4.7k ohm resistor have an arrow pointing to the middle of the 100 ohm resistor? I know the 100ohm is a varying resistor but if I were to actual solve the circuit for output and leave the varying resistor at 100ohms would the 4.7k be connected to node 3 or to node 1? And if it is indeed connected to node 3 then how would I go about doing nodal analysis at that node since I don't know the current to the left of the node, I just have a voltage source.

Correct if I'm wrong but at node 3 the nodal equation is: 5/x+(v3-v1)/4.7k+v3/100. Where x is the resistance, v3 is the voltage at node 3, and v1 is the voltage at node 6.
 
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Now are we assuming this op-amp is ideal? For laboratory purposes I would suspect that it is not--- however I did not read your lab to find out the Rin, A and Rout values.
 
Theelectricchild said:
Now are we assuming this op-amp is ideal? For laboratory purposes I would suspect that it is not--- however I did not read your lab to find out the Rin, A and Rout values.

Yes, it's an ideal op-amp.
 
Hi,

I forgot my nodal analysis learned during 1st year, but I can give you some light on the 100 ohms varying resistor.

Whether the varying resistor is 100 ohms or 1kohms is of no concern in the amplifier circuit because it effectively acts as a potentiometer OR potential divider. Whatever the voltage at node 2 depends on the position of the dial: If it is at node 3, the voltage at node 2 is the full 5V : If it is at node 1, the voltage at node 2 is the full 0V : If it is halfway in between, it is 1/2 * 5V = 2.5 V.
This is because as you turn the dial downwards, the resistance with respect to node 1 decreases, and the resistance with respect to node 3 increases.
The voltage at node 2 can be found using the potential divider formula:
R1/(R1+R2) * Vcc.
 

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