Voltage divider with added resistance

[jegues]

Homework Statement

See figure.

Homework Equations

N/A.

The Attempt at a Solution

First I did KVL in the large loop and the small loop so,

$$KVL_{large}: i_2R_1 + (i_1+i_2)R_2 - V_s = 0$$

$$KVL_{small}: i_1R_m - i_2R_1 = 0$$

Using KVL of the small loop one can say the following,

$$i_1 = \frac{i_2R_1}{R_m}$$

Plugging that into the KVL equation of the larger loop and simplify I found that,

$$i_2 = \frac{V_s}{R_1 + \frac{R_1R_2}{R_m} + R_2}$$

and for $$i_1$$ simply plug and chug.

Now this is a pretty messy equation for the current I2 which leads me to believe I've done something wrong... Is this the case or am I just worrying myself over nothing?

If so, what would be the better approach to solving this problem?

Thanks again!

 

[berkeman]

I personally prefer to use KCL equations, but that's up to personal preference.

Your equation looks very reasonable (I didn't check the detailed math). Look at the limits as Rm = 0 (short circuit) and Rm = infinity (open circuit). Your equation appears to give the right values. What about the test case when Rm = R1?

 

[jegues]

What about the test case when Rm = R1?

I'm not entirely sure what you're getting at here. If Rm = R1 then those resistors in parallel will become a R1/2 in series with R2, and we have ourselves another simple voltage divider.

How can I use this to further check my answer?

 

[cepheid]

I prefer using KCL as well. Look at the node that is at voltage V₂'. The current going into this node has to be equal to the current coming out of it. The current coming out of it is just V₂' / R₂. The current going into it is just I₁ + I₂, where I₂ = (V₁' - V₂') / R₁ and I₁ = (V₁' - V₂') / R_m. Simple.

I'm not entirely sure what you're getting at here. If Rm = R1 then those resistors in parallel will become a R1/2 in series with R2, and we have ourselves another simple voltage divider.

How can I use this to further check my answer?

You can use it to further check your answer by seeing if this simple voltage divider gives you the same answers for V₁' and V₂' as your equations for the more general case do.

 

[berkeman]

I'm not entirely sure what you're getting at here. If Rm = R1 then those resistors in parallel will become a R1/2 in series with R2, and we have ourselves another simple voltage divider.

How can I use this to further check my answer?

Only that I2 goes up, and by what looks like a reasonable amount. I didn't have any quantitative magic in mind. You could also probably just plug in some real values and see if the equation you've come up with gives the right answers. Or re-do the analysis with KCL -- that's different enough that it should show up any errors.

 

[jegues]

Or re-do the analysis with KCL -- that's different enough that it should show up any errors.

I'm going to give this a shot when I get home. I'll update this thread with my results.

I'd love it for you to come back and check my work again if possible!

 

[jegues]

Okay so using KCL,

$$\frac{V_2^'}{R_2} = \left( \frac{V_1^{'} - V_2^' }{R_1} \right) + \left( \frac{V_1^{'}- V_2^' }{R_m} \right)$$

I'm kinda lost as to what to do now, I need another equation right?

 

[cepheid]

Okay so using KCL,

$$\frac{V_2^'}{R_2} = \left( \frac{V_1^{'} - V_2^' }{R_1} \right) + \left( \frac{V_1^{'}- V_2^' }{R_m} \right)$$

I'm kinda lost as to what to do now, I need another equation right?

Do you? Don't you already know V₁'? Furthermore, I₁ and I₂ are known in terms of the two voltages. So how many unknowns does that leave?

 

[jegues]

Do you? Don't you already know V₁'? Furthermore, I₁ and I₂ are known in terms of the two voltages. So how many unknowns does that leave?

Hmm... Is V1 defined as follows,

$$V_{1}^{'} = V_{s} - V_{2}^{'}$$

?

If so, then I'd only have one unknown, $$V_{2}^{'}$$, and I could solve it!

 

[cepheid]

Hmm... Is V1 defined as follows,

$$V_{1}^{'} = V_{s} - V_{2}^{'}$$

?

If so, then I'd only have one unknown, $$V_{2}^{'}$$, and I could solve it!

Hmm? The node labelled V₁' and the + terminal of the voltage source are connected by an ideal wire, are they not? Therefore, they should be at the same electric potential.

 

[jegues]

Hmm? The node labelled V₁' and the + terminal of the voltage source are connected by an ideal wire, are they not? Therefore, they should be at the same electric potential.

Sorry, as you can tell this is where I have lots of my difficulties. I often get confused with node voltages.

So,

$$V_{1}^{'} = V_{s}$$

I'm so used to seeing voltage across an element, as soon as someone writes it on a node I get confused.

Any ideas for some good exercises/readings/tips/suggestions to help clarify things?

EDIT:

After thinking it over, that wouldn't make any sense either. If V1' = Vs then when we put KVL on the larger loop we'd find that V2' = 0 but there is current passing through the R2 so is must have some voltage across it.

What am I misunderstanding?

 

[cepheid]

Sorry, as you can tell this is where I have lots of my difficulties. I often get confused with node voltages.

So,

$$V_{1}^{'} = V_{s}$$

I'm so used to seeing voltage across an element, as soon as someone writes it on a node I get confused.

Any ideas for some good exercises/readings/tips/suggestions to help clarify things?

Indeed, a voltage is a difference between electric potentials, so I can see why labelling a single point as having a "voltage" might be confusing. The labelling of the node with V₁' is intended as a statement that the electric potential at that node is V₁'. But, you might protest that electric potential is kind of like altitude -- its value is meaningless unless if you specify a reference point, and the choice of reference point is arbitrary -- only differences in potential really matter. So what is meant by "the electric potential" at that point?

In circuits, by convention, all potentials are measured with respect to the point that is labelled as ground. In other words, ground is the reference point -- we define the electric potential to be 0 V there. Therefore, when we say that the "voltage" at that node is V₁', what we really mean is that the difference in electric potential between that node and ground is V₁'. Does that alleviate your confusion?

 

[cepheid]

EDIT:

After thinking it over, that wouldn't make any sense either. If V1' = Vs then when we put KVL on the larger loop we'd find that V2' = 0 but there is current passing through the R2 so is must have some voltage across it.

What am I misunderstanding?

How do you figure that, exactly?

 

[jegues]

Does that alleviate your confusion?

Your explanation definitely clears some confusion up with my understanding of node voltages, but I'm still kinda confused about what V1' is equal to. (See the EDIT in my post above).

When you say the following,

the "voltage" at that node is V1', what we really mean is that the difference in electric potential between that node and ground is V1'.

This makes me want to write the following voltage for V1',

$$V_{1} = V_{s} - V_{2}^{'}$$

Because these are the voltages across the two endpoints or nodes of our element, and one branches down to connect to ground. It feels sort of intuitive but I'm still not entirely convinced it's correct.(We determined it wasn't previously)

I might still be a little confused but maybe once I see how V1' is determined I'll have a better feel of what's going on.

 

[jegues]

How do you figure that, exactly?

$$KVL_{largeloop}: V_{1}^{'} + V_{2}^{'} - V_{S} = 0$$

If,

$$V_{1}^{'} = V_{S}$$

then,

$$V_{2}^{'} = 0$$

 

[cepheid]

This makes me want to write the following voltage for V1',

$$V_{1} = V_{s} - V_{2}^{'}$$

Because these are the voltages across the two endpoints or nodes of our element, and one branches down to connect to ground. It feels sort of intuitive but I'm still not entirely convinced it's correct.(We determined it wasn't previously)

I might still be a little confused but maybe once I see how V1' is determined I'll have a better feel of what's going on.

I see what you are trying to do, so let me explain further. Let's call the node that has been labelled V1', "node 1." The potential difference between node 1 and ground is V1' volts. In other words, if I take a voltmeter and touch one lead to node 1 and the other lead to ground, the voltmeter should read V1' volts.

Now, my claim is that this potential difference between node 1 and ground is equal to Vs. If that's true, then when I "travel" from ground to node 1, no matter what route I take through the circuit, I should gain an electric potential of Vs volts. How can I travel from ground to node 1? There are two possible routes. I could go through the branch that has the voltage source. If I do so, then clearly I WILL gain an electric potential of Vs volts, since I have moved across the voltage source. This serves to illustrate that the difference in potential between node 1 and ground is equal to the difference in electric potential between the terminals of the source (since the + terminal is connected to node 1 and the - terminal is connected to ground).

The second route I could take between ground and node 1 is across the branch with the two resistors R1 and R2 in it. When I move across this branch, I gain an electric potential equal to V_R2 + V_R1, where V_R2 is the voltage across R2, and V_R1 is the voltage across R1. Agreed? We know what these voltages across the resistors are:

$$V_{R2} = V_2^\prime - 0$$

$$V_{R1} = V_1^\prime - V_2^\prime$$

Agreed?

Therefore, the total electric potential I gain by going from ground to node 1 by traveling across the two resistors is just:

$$V_{R1} + V_{R2} = (V_1^\prime - V_2^\prime) + (V_2^\prime) = V_1^\prime$$

So in summary, going across the branch with the source in it, I gain an electric potential of Vs. Going across the branch with the two resistors in it, I gain an electric potential of V_R1 + V_R2 = V1'. Since both routes involve traveling between the same two points, the potential difference must be the same for both routes, meaning that it must be true that:

$$V_s = V_{R1} + V_{R2} = V_1^\prime$$

Do you see why this equation is right and the one you wrote down is not?

 

[cepheid]

$$KVL_{largeloop}: V_{1}^{'} + V_{2}^{'} - V_{S} = 0$$

This KVL equation is simply wrong, because the voltage across the right side of the loop is not V1 + V2. As I explained in my previous post, KVL demands that the voltage across the source is equal to the sum of the voltages across the two resistors R1 and R2, so that the KVL equation should read:

V_s = V_R1 + V_R2

I won't elaborate further, since I would just be rehashing my previous post.

 

[jegues]

I see what you are trying to do, so let me explain further. Let's call the node that has been labelled V1', "node 1." The potential difference between node 1 and ground is V1' volts. In other words, if I take a voltmeter and touch one lead to node 1 and the other lead to ground, the voltmeter should read V1' volts.

Now, my claim is that this potential difference between node 1 and ground is equal to Vs. If that's true, then when I "travel" from ground to node 1, no matter what route I take through the circuit, I should gain an electric potential of Vs volts. How can I travel from ground to node 1? There are two possible routes. I could go through the branch that has the voltage source. If I do so, then clearly I WILL gain an electric potential of Vs volts, since I have moved across the voltage source. This serves to illustrate that the difference in potential between node 1 and ground is equal to the difference in electric potential between the terminals of the source (since the + terminal is connected to node 1 and the - terminal is connected to ground).

The second route I could take between ground and node 1 is across the branch with the two resistors R1 and R2 in it. When I move across this branch, I gain an electric potential equal to V_R2 + V_R1, where V_R2 is the voltage across R2, and V_R1 is the voltage across R1. Agreed? We know what these voltages across the resistors are:

$$V_{R2} = V_2^\prime - 0$$

$$V_{R1} = V_1^\prime - V_2^\prime$$

Agreed?

Therefore, the total electric potential I gain by going from ground to node 1 by traveling across the two resistors is just:

$$V_{R1} + V_{R2} = (V_1^\prime - V_2^\prime) + (V_2^\prime) = V_1^\prime$$

So in summary, going across the branch with the source in it, I gain an electric potential of Vs. Going across the branch with the two resistors in it, I gain an electric potential of V_R1 + V_R2 = V1'. Since both routes involve traveling between the same two points, the potential difference must be the same for both routes, meaning that it must be true that:

$$V_s = V_{R1} + V_{R2} = V_1^\prime$$

Do you see why this equation is right and the one you wrote down is not?

Thank you for the excellent post cepheid this clears up a lot of confusions I had concerning node voltage. I can even refer back to this post if I ever get confused again!

Now back to the original question.

Since we determined that,

$$V_{1}^{'} = V_{s}$$

Then we should find that,

$$\frac{V_2^'}{R_2} = \left( \frac{V_s^{'} - V_2^' }{R_1} \right) + \left( \frac{V_s^{'}- V_2^' }{R_m} \right)$$

Simplifying I find that,

$$V_{2}^{'} = \frac{(\frac{1}{R_1} + \frac{1}{R_m})V_{s}}{1 + (\frac{1}{R_{1}} + \frac{1}{R_m})}$$

EDIT: The part on the right of the fraction line should be the denominator in the fraction.

How's this look?