# Op Amp Requivalent Please help!

[SOLVED] Op Amp Requivalent Please help!

1. Homework Statement
I'm working on Ideal op amps and I need to find the R equivalent of the system.
I'm not sure if this term is used in other places but R equivalent = total Resistance of the circuit.

http://img169.imageshack.us/img169/4386/reqrk6.png [Broken]

2. Homework Equations

3. The Attempt at a Solution
I have to place a voltage at the circle at the top left and use the voltage to divide the current to get the Req, but I don't know how to find the current. I think I have to use a current divider at the 10 and 40 k ohms but I can't seem to get the answer of 14 k ohms as the answer in the back of the book. Please help!

I'm also confused if this is the Negative Resistance mentioned in the book. But the given equation for Req = -(R1/R2)R

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## Answers and Replies

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The Electrician
Gold Member
You say "I think I have to use a current divider at the 10 and 40 k ohms...", but I don't see a 10k resistor.

However, I changed the very first resistor to the right of the little circle to 10k, and I got 14k for Req. Here's how I did it:

Apply 1 volt to the little circle. The bottom branch of 40k and 30k form a voltage divider. The voltage at the + input of the opamp will be 3/7 volt, and if the opamp has infinite gain, the voltage at the - input will also be 3/7 volt. This means the voltage across the 40k resistor in the bottom branch will be 4/7 volt, and the voltage across the 10k in the top branch will also be 4/7 volt. The current in the 40k in the bottom branch will be (4/7)/40000 and the current in the 10k in the top branch will be (4/7)/10000. The total current drawn from the applied 1 volt will be (4/7)/40000 + (4/7)/10000 = 7.142857E-5 amps. Then Req = 1 volt / 7.142857E-5 = 14000 ohms.

ohh I see! I think you're right, I misread it when I drew that on paint, the top resistor is 10k ohms. But one thing I didn't quite get was why the voltage across the 10k resistor is also 4/7. Thanks for helping.

The Electrician
Gold Member
When you're solving opamp problems, unless the problem says otherwise, they almost always assume the opamp is ideal: it has infinite gain and the inputs don't draw any current (or negligible current at most). This means that if the opamp is operating normally and the output isn't driven to the power supply rails (which is usually the case if the negative feedback is greater than the positive feedback), then the + and - inputs will be at the same voltage (will have no voltage difference between them). These two assumptions will allow you to easily solve a lot of opamp problems.

Thank you.