Solving a Puzzling Circuit: Find V2

[kaspis245]

Homework Statement

Three equal resistors and three equal voltmeters are connected as shown. V₁ shows 10V and V₃ shows 8V. Find V₂.

Homework Equations

Ohm's Law

The Attempt at a Solution

I don't know how to solve it since the voltmeters are connected in parallel. Shouldn't they show the same?

 

[SammyS]

Homework Statement

Three equal resistors and three equal voltmeters are connected as shown. V₁ shows 10Ω and V₃ shows 8Ω. Find V₂.

Homework Equations

Ohm's Law

The Attempt at a Solution

I don't know how to solve it since the voltmeters are connected in parallel. Shouldn't they show the same?

Just how is it that the voltmeter is showing a measurement in Ohms ?

 

[kaspis245]

Sorry, I've corrected it.

 

[gneill]

If the voltmeters shown were ideal (infinite impedance) then there would be no complete path (circuit) for current to flow, thus no potential drops could occur across the resistors and they'd all show the same reading. But we are told that there are potential drops since the two given voltage readings are different. Does that suggest anything to you?

 

[kaspis245]

If the voltmeters shown were ideal (infinite impedance) then there would be no complete path (circuit) for current to flow, thus no potential drops could occur across the resistors and they'd all show the same reading. But we are told that there are potential drops since the two given voltage readings are different. Does that suggest anything to you?

Nope. Are you saying that voltmeters have their own resistance?

 

[gneill]

Nope. Are you saying that voltmeters have their own resistance?

Yes. They must if there's current flowing. The only path back to the source E is via voltmeters.

 

[kaspis245]

So should I use Kirchhoff's rule or what? Please help, I am new to all of this.

 

[gneill]

Well you'll surely need Kirchhoff simply to write circuit equations. I can't think of a standard approach for this problem since simply writing node or loop equations will leave you with too many unknowns. That leaves being clever and getting creative

My suggestion would be to reduce the number of unknowns by either combining them or making them irrelevant. How does one do that, you ask? Well for starters you've got only two resistor values (Let's call the voltmeter resistance r) in the circuit, and something tells me from the repeating blocks (Resistor + Voltmeter sections) that maybe the ratio of the r to R could be important. So make r = n*R.

Next, the source E is an unknown. Let's get rid of it! Chop it off along with the first R and replace the first voltmeter with a 10V source. Note that this won't affect the rest of the circuit to the right of it! v1 is now fixed at 10V just as before and the "load" on it is the same.

See if you can then write a node equations to solve for v2, keeping an eye out for maybe eliminating either R or n along the way. Remember that you have the voltage at the output of the voltage divider formed by v2 and resistors R and nR (a voltmeter).

 

[kaspis245]

Here's what I did:
r - resistance of the voltmeter

Obviously, I can't solve it using Kirchhoff's rule.

 

[gneill]

Obviously, I can't solve it using Kirchhoff's rule.

Is that the only way Kirchhoff can be applied?

Hint: Let r = n*R, and use nodal analysis to find an expression for v2 (v2 is the only essential node in this reduced circuit). Look to cancel R's and leave you with an expression for v2 in terms of n. You also happen to know that the V3 branch is a voltage divider producing 8V from v2...

 

[litup]

The problem doesn't say what the resistance of the voltmeters are, typical DVM's come in around 10 megohms which would be close enough to infinity to ignore current flow through the meter, assuming the resistor values are low, 100 ohms or so. They don't specify the battery voltage so it looks like you have to solve for that also. The thing I don't see here is how the voltmeters can read different voltages. For instance, we make the battery 30 volts, the resistors each 10 ohms, than there would be 1 amp flowing throught the series resistances and 10 volts measured across each resistor assuming infinite or close to infinite DVM resistance.

That would be the voltage you would read across each resistor assuming infinite DVM resistance, but they specify 10 and 8 ohms, so if you treat it like that you would have a network of 6 resistors and the center one designated V2 would have an unknown resistance value but V3 at 8 ohms would be in series with the 10 ohms I picked or 18 ohms.

I think they need to specify exactly what part is what, if the V1,V2, and V3 really is showing those ohm numbers. Even an analog meter would be 1000 ohms minimum, some 50,000 ohms so they need to specify the problem better. Are the 'ohms' attached to the meters actually ohms?

Just looking at the numbers given I would say V2 should be 9 ohms but that is just a guess, taking the problem at face value. It would certainly help if they specified a battery voltage, at least you could figure out the current through the equal resistors, instead of just making up a voltage.

 

[gneill]

The problem doesn't say what the resistance of the voltmeters are, typical DVM's come in around 10 megohms which would be close enough to infinity to ignore current flow through the meter, assuming the resistor values are low, 100 ohms or so. They don't specify the battery voltage so it looks like you have to solve for that also. The thing I don't see here is how the voltmeters can read different voltages. For instance, we make the battery 30 volts, the resistors each 10 ohms, than there would be 1 amp flowing throught the series resistances and 10 volts measured across each resistor assuming infinite or close to infinite DVM resistance.

That would be the voltage you would read across each resistor assuming infinite DVM resistance, but they specify 10 and 8 ohms, so if you treat it like that you would have a network of 6 resistors and the center one designated V2 would have an unknown resistance value but V3 at 8 ohms would be in series with the 10 ohms I picked or 18 ohms.

I think they need to specify exactly what part is what, if the V1,V2, and V3 really is showing those ohm numbers. Even an analog meter would be 1000 ohms minimum, some 50,000 ohms so they need to specify the problem better. Are the 'ohms' attached to the meters actually ohms?

Just looking at the numbers given I would say V2 should be 9 ohms but that is just a guess, taking the problem at face value. It would certainly help if they specified a battery voltage, at least you could figure out the current through the equal resistors, instead of just making up a voltage.

Voltmeters do not read in Ohms. They read in Volts. The original problem statement had a typo there, and was subsequently corrected.

Clearly this is not a "real world" circuit with "real world" voltmeters. It's a circuit puzzle problem. The resistance of the voltmeters is an unknown, but we know that they are identical. There is enough information given to solve the puzzle.

If you wish, replace the voltmeters with resistors of value r, and specify that two known node voltages are 10 and 8 volts.

 

[kaspis245]

Using nodal analysis:

Current through R₁: 10/R
Current through R₂: V₂/R
Current through V₃: 8/r
Current through V₂: V₂/r

I=(10+V2)\R+(8+V2)\r

I don't know where to use that r=nR.

 

[NascentOxygen]

The current through R2 is the potential difference between its terminals divided by R. What is the voltage at each end of that resistor?

 

[gneill]

Here's the "reduced" circuit:

Ignore the 8V for now: You can use it later to form another equation. Write the node equation, treating the right branch as R + nR. Note that the current through the leftmost resistor is NOT 10/R. It's (10 - v2)/R if you choose to sum currents flowing into the node, or (v2 - 10)/R if you choose to sum currents flowing out of the node. You need to use potential differences across each resistor.

 

[kaspis245]

You mean that I should treat the right branch like this: I₁ = V₃/(R+nR)

 

[gneill]

You mean that I should treat the right branch like this: I₁ = V₃/(R+nR)

No, it's V₂ at the node. The total branch resistance is correct.

 

[kaspis245]

So I get this:

I = I₁+I₂ = V₂/(R+nR) + V₂/nR

 

[gneill]

So I get this:

I = I₁+I₂ = V₂/(R+nR) + V₂/nR

You don't want I, you want an expression for v2. You've left out the branch with the voltage source. Your node equation should contain only voltage sources and resistances, no "I".

 

[kaspis245]

Well, I could find the total resistance of the current and divide 10V by it, but I would get a very nasty expression. I don't think this would work.

 

[gneill]

Well, I could find the total resistance of the current and divide 10V by it, but I would get a very nasty expression. I don't think this would work.

You don't need the current! You need the node equation for v2.

 

[kaspis245]

How about this:

$\frac{10}{R}-\frac{V_2}{nR}-\frac{V_2}{(R+nR)}=0$

 

[NascentOxygen]

How about this:

$\frac{10}{R}-\frac{V_2}{nR}-\frac{V_2}{(R+nR)}=0$

Getting closer!

If you explain in words what each term represents, you'll probably realize your mistake.

 

[kaspis245]

Is this correct:

$\frac{10-V_2}{R}-\frac{V_2}{nR}-\frac{V_2}{(R+nR)}=0$

 

[NascentOxygen]

Is this correct:

$\frac{10-V_2}{R}-\frac{V_2}{nR}-\frac{V_2}{(R+nR)}=0$

That looks right.

 

[kaspis245]

I get this:

$V_2= \frac{10n+10n^2}{1+n+n^2}$

I don't see how this would help me.

 

[NascentOxygen]

I get this:

$V_2= \frac{10n+10n^2}{1+n+n^2}$

I don't see how this would help me.

You were asked to find V₂, and that's what you have. (n is a constant, though undetermined yet.)

I don't get the denominator that you have. One of us must be in error.

 

[kaspis245]

Sorry, the answer is:

$V_2 = \frac{10n+10n^2}{1+3n+n^2}$

What should I do with this?

 

[NascentOxygen]

Can you find V₃ in terms of V₂ and equate this to 8?

 

[kaspis245]

Can you find V3 in terms of V2 and equate this to 8?

I don't understand this. You mean I should write it like this:

$\frac{10-V_2}{V_2}-\frac{V_2}{nR}-\frac{V_2}{R}-\frac{8}{nR}=0$

 

[NascentOxygen]

I had in mind the potential divider R & nR that produces V₃ from V₂. This allows you to express V₃ in terms of V₂ and n.

I'm not sure what gneill had in mind, but once you have calculated n I think you'll be as good as done.

Do you happen to know the correct answer?

 

[kaspis245]

I don't know the answer.

At this point I am completely confused. Appreciate if you just write the solution. It would be much easier for me to understand. Is this correct:

$V_3=\frac{nR}{R+nR}⋅V_2$

 

[NascentOxygen]

$\frac{nR}{R+nR}⋅V_2 = 8$

Now substitute your earlier expression for V₂ and solve for the only unknown, n. A few things should cancel, making it easier than it looks.

 

[kaspis245]

I get n=13.232. What now?

 

[kaspis245]

I placed n into this equation:

$V_2= \frac{10n+10n^2}{1+n+n^2}$

and got V₂=8.73 V

Is this correct?

 

[gneill]

I get n=13.232. What now?

That's a bit off from what I'm seeing. But you've made excellent progress!

Myself, I didn't solve for n first, but rather substituted $n = \frac{8}{v2 - 8}$ from the voltage divider part into the node equation which then reduced to a pretty nice quadratic for v2.

 

[kaspis245]

Sorry, I've recalculated it and got this equation:

$n^3-11n^2-16n-4=0$

Then let wolfram alpha do the work and got n=12.325

This way V₂=9.94V

Is it correct?

 

[gneill]

The value for n looks good. The value of V2 does not. What expression did you use?

 

[kaspis245]

I used this expression:

$V_2= \frac{10n+10n^2}{1+3n+n^2}$

which is derived from:

$\frac{10-V_2}{V_2}-\frac{V_2}{nR}-\frac{V_2}{R+nR}=0$

 

[gneill]

I used this expression:

$V_2= \frac{10n+10n^2}{1+3n+n^2}$

which is derived from:

$\frac{10-V_2}{V_2}-\frac{V_2}{nR}-\frac{V_2}{R+nR}=0$

I presume that first term should be over R, not $V_2$.
Okay, that should work. But I'm seeing a different value for $V_2$ using that equation.

I would have used the simpler voltage divider derived equation myself.

 

[kaspis245]

I am such an idiot, I've made another mistake in my calculations.

Here's my final answer:

$V_2=8.65 V$

 

[NascentOxygen]

Sorry, I've recalculated it and got this equation:

$n^3-11n^2-16n-4=0$

Then let wolfram alpha do the work and got n=12.325

This way V₂=9.94V

Is it correct?

A cubic?? You have probably overlooked a term in the numerator which would have canceled a factor of this, leaving you with just a quadratic to solve.

n=12.325 is my answer, too, from the quadratic solution

 

[kaspis245]

As long as we are getting the same answer I am ok with that