Is that how it works basically?
I think that is correct. Wherever the slide is splits the total resistance in two
That is exactly how it works. The two resistor values sum to the total resistance of the potentiometer.
It's like cutting a piece of string: You can cut it anywhere but the lengths of the two pieces still add up to the length of the whole string.
Please see my comments in Post #4 of this thread:
The potentiometer in the drawing in the OP of this thread is being used inappropriately, and the circuit will have reliability problems.
Berkeman -- is it just because I didn't do p/2 ? Why would 2 people said I split it correctly, though?
If I may carry on with the exercise....I need to find the value at point a IF the potentiometer is at 25% of its value
Did I get it right?
Hello again Femme_physics! You say 25%, but in your working you've set it at the 50% mark. Which should it be?
Oops, I forgot to divide by two. But the result of Va = 2 [V] still ends up the same :)
And then Va = Vb due to the fact it's a feedback op-amp
Now I imagine this is an inverter op-amp, ergo
Vc = Vout = -8 [V]
Who's the queen? :)
When a is 25%, the 8K potentiometer is divided into two sections. One is 25% of 8K and the other is 75% of 8K. End-to-end the potentiometer still totals 8K.
So you have, for example, RP1=0.75P and RP2=0.25P, and Vi=4v, and you still think Va=2V?? What happened to our "potential divider" expert from two weeks ago?
The crown has yet to be claimed. ♔
Wait, gneill, I'll reiterate the question word by word
Potentiometer P is set at 25% of its value (the resistance between point a and earth is 25% of P value).
I'm just all the more confused now. If I split the potentiometer to 2 resistors, don't they both must equal to the value of I set the potentiometer at? According to the question IMO it's 2 volts TOTAL. i.e. their sum is 2 volts and they split equally...the way I read it
NasOx-- just saw ur reply give me a moment :)
...so the portion above point a must be 75% of P value.
The physical implementation of a potentiometer has a resistive strip whose total end-to-end resistance is the "value" of the potentiometer. Each end of the strip has a contact. There is a third contact that can slide from end to end along the surface of the strip. This contact "divides" the strip into two portions according to the slider location. The resistance between the sliding contact and an end of the strip is proportional to the length of strip lying between the contact point and that end. The portions can be any ratio of lengths (and thus resistances) but their sum must still add up to the net length (resistance) of the strip.
From the original circuit diagram it looks like the potentiometer can "tap" any value between 0 and 4V.
She didn't have to deal with weird potentiometer thingies!!
ahh...I see! So...
No. Let's go back to your diagram in post #5 of this thread. See on RP1 where you wrote "p/2"? Replace that writing with 3p/4. See RP2? Replace "p/2" there with p/4.
That's the only change. Now analyze it. https://www.physicsforums.com/images/icons/icon6.gif [Broken]
So P in that case is still its original value. It's only from point a to the ground that the 0.25 is?
Fine, but how was I supposed to have known all that by the question if it hasn't been defined like that?
The connection is 25% along the length of the resistor element (i.e., measured from the earthed end). So now the op-amp's (+) input is being supplied 25% of the 4 volts.
The voltage divider nature of the potentiometer is correct. It is bad design practice to connect the wiper of a potentiometer directly to just an opamp input. The reason is described in the post that I linked to.
Ahhh perfect explanation and diagram. See, I thought that a potentiometer is a device that regulate the amount of resistance of a resistor. But it just regulates the spread of differences....makes sense.
Yes sir! I do most certainly can prove it with calculations, too!
No, not quite there yet. ☹
Does it not seem strange that you arrived at your answer without involving Va anywhere in the calculation? You went to all the trouble of determining what the potentiometer setting meant, then calculated the voltage at the non-inverting input of the op-amp, then calmly discarded all that work and went back to the Rf/Rin equation you remembered from a simpler circuit previously
I'm afraid you don't get out of it that easily.
So let's go back to your step ➁...
Va=Vb is correct, and as always this relationship applies for any op-amp functioning as an amplifier.
You determined Va = 1V. Therefore, we can say Vb = 1V
Now, let's look at the network connected to the op-amps (–) input. There's Vi at one end of R1 and Vb at its other end. So what is the expression for current through R1? Use Ohm's Law.
Next, look at R2. What is the voltage at each of its ends? So determine an expression for the current through R2.
The op-amp, as always, draws practically no current into either of its input terminals, so we are left with the customary relation, current in R1 = current in R2. So equate your expressions for these currents, and solve for Vout.
See how you go this time.
I guess I just didn't think they are related...hmm...
That's usually how it turns out :)
Ir1 = Va/R1 = 1/3000 = 0.333 [mA]
The voltage at R2 must be 3 [V]...as 4-1 = 3[V]
Ir2 = Vr2 / R2 = 0.5 [mA]
Same result for when using the formula!
Nope. What's the potential drop across R1? It's not Va --- What happened to Vin? Vin is at one end of R1 and Va at the other, so the potential difference is...
Separate names with a comma.