negativepotentiometerresistorsvoltage divider
I have the following voltage divider circuit with a potentiomenter, and I need to find the value of the resistors.
simulate this circuit – Schematic created using CircuitLab
How can I find the values of R1 and R3?
I believe both resistors could have the same value. That would be nice, if that is the case.
Voltage Divider Basics and Analysis
I noticed on a later comment that you were really trying to understand how this could (potentially) be done using voltage divider theory. To start, let me remind you what a voltage divider really is:
Take this resistive voltage divider from Wikipedia:
The common equation to find the output is: V_out = R2 / (R1 + R2) * V_in
But what this really boils down to is this: V_out = R2 * I_R2
Of course, in this circuit, the current through R1 and R2 is equal, so finding the value is easily done by dividing the total voltage by the total current: I = V_in / (R1 + R2)
Because R2 is located between V_out and ground, the voltage across R2 is equal to V_out. However, what would happen if a third resistor were placed in parallel to R2? Consider this modified version of the above image:
The current relationship has now changed. In this circuit, I_R1 = I_R2 + I_R3. This makes the equation for the output a bit more complex: V_out = (R2 || R3) / (R1 + (R2 || R3)) * V_in
Because R2 and R3 are in parallel: (R2 || R3) = (R2 * R3) / (R2 + R3)
But the voltage across R2 is still equal to V_out. Adding another series resistor changes this:
V_out = ((R2 || R3) + R4) / (R1 + (R2 || R3) + R4) * Vin
Now, the voltage across R2 is really equal to V_out - V_R4. Adding more resistors will only further skew the relationship between V_out and V_R2. Adding a resistor in series to R3 doesn't change things too much:
V_out = ((R2 || (R3 + R4)) + R5) / (R1 + (R2 || (R3 + R4)) + R5) * Vin
where R2 is in parallel with the series combination of R3 and R4... This is still a big resistive voltage divider with random parallel and series resistors. The voltage across R2 is equal to V_out - V_R5.
The real complication comes with the introduction of the bottom half of the Wheatstone Bridge formed by resistors 2 through 6. Your particular circuit is very simple because all of the resistors are of equal value (the bridge is balanced), but it is a bit harder to formulate equations for a generic circuit which might have a voltage across R4:
If you really want to keep using basic resistive relationships, the next best step is to simplify the circuit. One method is to do a Delta-Wye Transform, as you mentioned in the question. However, it makes a lot more sense to do it on the the bottom half or right side of the bridge where R2 is not involved. For example, converting the bottom half of the bridge yields:
The new resistor values are found as:
But this still doesn't solve the problem because you need to know the voltage across R2 which is now equal to V_out - V_Na. This can still be solved using voltage dividers if you use a couple of different points as outputs:
The final solution (the voltage across R2) can finally be found using a combination of various voltage dividers and the following equations:
Since you said you already found the answer using Mesh Analysis, I will go ahead and put the values you should get from the above equations:
So it is possible to solve the circuit using nothing but voltage dividers, but we needed help from a resistor transformation. All of the basic circuit analysis techniques are based on Ohm's law: V = I*R, some of them just make more sense to use at times than others do.
Using Nodal Analysis
Considering all of your resistors are the same value, there are a few tricks to easily reduce the entire circuit down, but since that is pretty unrealistic in real life, I think it would be better to teach you a real analysis tool. There are many ways to analyse a circuit, but one of the easiest and best for many types of problems is Nodal Analysis. It is very easy to learn.
1) According to Kirchhoff's current law (KCL), the current going into any one node (point in the circuit) must be equal to the current coming out of any one node. So divide the circuit up into individual currents flowing from one node to another. The direction of the current flow is entirely up to you, it will just affect the polarity of the resulting current:
2) Form relationships between the currents:
3) Express the currents as voltages/resistances
4) Using Algebra, combine and reduce the equations to find the values of the unknown voltages. To answer your specific question, the voltage across R2 would be V3 - V6 which is equal to R2 * I2.
The idea of output resistance comes from a thought where you place the circuit you have into a black box where you cannot see it and don't know what it is, anymore. You have access to two terminals from this box. You can measure the voltage with a voltmeter and that tells you something. You can then short out the terminals using an ammeter and see what current you measure (hopefully, this doesn't damage whatever is in the black box or you when trying to do this step.) You can then mentally propose that the circuit in the black box is the equivalent of this:
simulate this circuit – Schematic created using CircuitLab
Since using a perfect voltmeter means that there isn't any current used during the measurement, the voltage you measured across A to B with the voltmeter tells you the voltage of this hidden equivalent voltage source. (The resistor doesn't drop a voltage without current flowing through it, so the measurement accurately reflects that hidden source.) Now, when you short A to B using an ammeter and measure the current, you can work out the effective resistance of R. It's just:
$$R=\frac{V_{voltmeter}}{I_{ammeter}}$$
That has to be the case, right?
That value you computed is the effective "output resistance" of this black box.
Now, let's return to your circuit and try to figure out what a black box measuring process would give us. Let's call the top resistor \$R_1\$ instead of just R and the bottom resistor \$R_2\$ also instead of just R. That way we can work this out, more generally.
If you just put a voltmeter up to \$V_{mid}\$ and ground, you'd measure a voltage:
$$V_{voltmeter} = V_{dd}\cdot\frac{R_2}{R_1+R_2}$$
And if you hooked up an ammeter up to \$V_{mid}\$ and ground, you'd measure a current:
$$I_{ammeter} = \frac{V_{dd}}{R_1}$$
Now, if you divide these, you get the following resistance:
$$R=\frac{V_{dd}\cdot\frac{R_2}{R_1+R_2}}{\frac{V_{dd}}{R_1}}=\frac{R_1 R_2}{R_1+R_2}$$
But that's just the parallel equivalent of the two resistors! In your case, where the values are both the same, this just means half the value (like your teacher said.)
So you can compute this black box resistor's equivalent value and, in fact, the circuit will behave as if that resistor value really were present. It's not, because you can see the actual circuit. But if you couldn't see the hidden circuit, you'd have to postulate this resistance being there.
There's a deeper, and for some more satisfying, approach using differential equations. But I'll avoid going there.
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