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

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The discussion focuses on how a varying resistor impacts the voltage at node 2 in an op-amp circuit. The 4.7k ohm resistor connects to node 3, and the varying 100 ohm resistor acts as a potentiometer, affecting the voltage at node 2 based on its position. When fully turned towards node 3, node 2 receives 5V; when turned towards node 1, it receives 0V, and halfway it receives 2.5V. The voltage at node 2 can be calculated using the potential divider formula. Understanding these relationships is crucial for solving the circuit effectively.
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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.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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