Reading of the voltmeter with closed switch

[moenste]

Homework Statement

(a) In the circuit shown below E is a cell of source (internal) resistance r and the resistance of R is 4.0 Ω. With the switch S open, the high resistance voltmeter reads 10.0 V and with S closed the voltmeter reads 8.0 V. Show that r = 1.0 Ω.

(b) If R were replaced by a cell of e. m. f. 4.0 V and source resistance 1.0 Ω with its negative terminal connected to B, what would be the reading of the voltmeter with S closed?

Answer: (b) 7.0 V.

2. The attempt at a solution
(a) E = I (r + R) → 10 = I (1 + 4) → I = 2 A. Then we find E = I r + V → E - V = I r → 10 - 8 = 2 r → r = 1 Ω. Looks correct.

(b) Maybe something like 10 - 4 = 1 I₃ + 1 I₁, 10 - V = 1 I₁?

 

[gneill]

(a) E = I (r + R) → 10 = I (1 + 4) → I = 2 A. Then we find E = I r + V → E - V = I r → 10 - 8 = 2 r → r = 1 Ω. Looks correct.

Yes, that looks okay.

(b) Maybe something like 10 - 4 = 1 I3 + 1 I1, 10 - V = 1 I1?

There's only one current circulating in the loop, so no need to introduce more current variables.

Have you learned about Kirchhoff's Voltage Law (KVL)? It would really help you to write these equations since it can be done in a formulaic manner. It would also help you to find these voltages in a methodical way.

 

[moenste]

Have you learned about Kirchhoff's Voltage Law (KVL)? It would really help you to write these equations since it can be done in a formulaic manner. It would also help you to find these voltages in a methodical way.

In my book I have Kirchhoff's rule 1: the algebraic sum of the currents flowing into a junction is zero, i. e. Σ I = 0, I₁ = I₂ + I₃.

And rule 2: in any closed loop, the algebraic sum of the EMFs is equal to the algebraic sum of the products of current and resistance, i. e. Σ E = Σ I R. Using this rule each resistor within a particular loop must be traversed in the same sense (either clockwise or anticlockwise).

KVL is a different name for the second rule?

 

[gneill]

In my book I have Kirchhoff's rule 1: the algebraic sum of the currents flowing into a junction is zero, i. e. Σ I = 0, I₁ = I₂ + I₃.

And rule 2: in any closed loop, the algebraic sum of the EMFs is equal to the algebraic sum of the products of current and resistance, i. e. Σ E = Σ I R. Using this rule each resistor within a particular loop must be traversed in the same sense (either clockwise or anticlockwise).

KVL is a different name for the second rule?

Yes, KVL is the second rule. In practical terms the rule states that the sum of the changes in PD that occur when traversing the loop is zero. So if you "walk" around the loop in either direction, summing up any rises and falls of PD on the components as you go, when you arrive back at your starting location that sum must be zero.

It's analogous to walking a closed path on a terrain and counting the changes in elevation. You expect the net change to be zero when you get back to the starting point.

 

[moenste]

Yes, KVL is the second rule. In practical terms the rule states that the sum of the changes in PD that occur when traversing the loop is zero. So if you "walk" around the loop in either direction, summing up any rises and falls of PD on the components as you go, when you arrive back at your starting location that sum must be zero.

It's analogous to walking a closed path on a terrain and counting the changes in elevation. You expect the net change to be zero when you get back to the starting point.

In our case it would be -4 V + 8 V = 0?

 

[gneill]

In our case it would be -4 V + 8 V = 0?

No. First of all it makes no mathematical sense, since 8-4 is clearly not zero. Secondly, you haven't accounted for all the potential changes. When I look at the circuit for part b I see two EMFs of 10 V, one of 10 V and one of 4 V, and two internal resistances of 1 Ohm. I don't know the current yet, but I can write the KVL sum around the loop using a variable $I$ for that current.

So, write the KVL equation for the loop with assumed current $I$. The sum, since your'e arriving back where you started having completed a closed loop, will be zero.

[edit: Fixed the EMFs. Too many threads ongoing to keep track! ]

 

[moenste]

No. First of all it makes no mathematical sense, since 8-4 is clearly not zero. Secondly, you haven't accounted for all the potential changes. When I look at the circuit for part b I see two EMFs of 10 V and two internal resistances of 1 Ohm. I don't know the current yet, but I can write the KVL sum around the loop using a variable $I$ for that current.

So, write the KVL equation for the loop with assumed current $I$. The sum, since your'e arriving back where you started having completed a closed loop, will be zero.

8 - 1 I - 4 - 1 I = 0
4 - 2 I = 0
I = 2 A

 

[gneill]

8 - 1 I - 4 - 1 I = 0
4 - 2 I = 0
I = 2 A

That's the idea, only one of the EMF's is 10 V, not 8 V.

 

[moenste]

That's the idea, only one of the EMF's is 10 V, not 8 V.

It's 10 when it's open. We have a situation when it's closed in (b).

 

[gneill]

It's 10 when it's open. We have a situation when it's closed in (b).

The EMF is inviolate. It doesn't change. It's the voltage drop across the internal resistance that makes the PD drop at the battery terminals. You measure the EMF of a battery by insuring that no current is flowing --- put a good voltmeter across the battery terminals with no other load. That was the purpose of part (a), to determine the EMF of the battery. It is 10 V. It stays 10 V no matter what.

 

[moenste]

The EMF is inviolate. It doesn't change. It's the voltage drop across the internal resistance that makes the PD drop at the battery terminals. You measure the EMF of a battery by insuring that no current is flowing --- put a good voltmeter across the battery terminals with no other load. That was the purpose of part (a), to determine the EMF of the battery. It is 10 V. It stays 10 V no matter what.

10 - 1 I - 4 - 1 I = 0
6 - 2 I = 0
I = 3 A

V = I R = 3 * (1 + 1) = 6 V?

 

[gneill]

10 - 1 I - 4 - 1 I = 0
6 - 2 I = 0
I = 3 A

V = I R = 3 * (1 + 1) = 6 V?

The current value looks good.

What are you calculating with that last line?

 

[moenste]

What are you calculating with that last line?

Voltmeter voltage (b) part.

 

[gneill]

Voltmeter voltage (b) part.

Then your calculation is not correct. The voltmeter is connected across EMFs and resistors. You've only assumed resistors and that they are connected in series between A and B. They are not.

Instead, do a KVL walk from A to B. You can choose either direction (via the upper path through the 10 V EMF, or the lower path via the 4 V EMF).

 

[moenste]

Then your calculation is not correct. The voltmeter is connected across EMFs and resistors. You've only assumed resistors and that they are connected in series between A and B. They are not.

Instead, do a KVL walk from A to B. You can choose either direction (via the upper path through the 10 V EMF, or the lower path via the 4 V EMF).

10 V - 3 A * 1 Ω = V
V = 7 V

I used 10 V as you said, not 8 V.

4 + 3 * 1 = V
V = 7 V

Honestly, I'm still struggling to understand where a plus or a minus should go. For example why is it not 4 - 3 * 1 = V? I know we can just play with the numbers to get my answer 10 - 3 * 1 = V and 4 + 3 * 1 = V and get V = 7 V in both formulas but I want to understand the logic behind : (.

 

[gneill]

Honestly, I'm still struggling to understand where a plus or a minus should go. For example why is it not 4 - 3 * 1 = V? I know we can just play with the numbers to get my answer 10 - 3 * 1 = V and 4 + 3 * 1 = V and get V = 7 V in both formulas but I want to understand the logic behind : (.

Make a drawing of the circuit. Draw in the current direction. For every resistor that the current passes through place a small + sign where it enters and a - sign where it exits the resistor. That's it. When you do your walk along a path of the circuit just account for the change in potentials using those signs. + to - is a drop so it's negative. - to + is a rise so that's positive.

Some people sum the drops and treat them as positive, while rises are accounted as negative. There's aesthetic reasons for this having to do with making the math look nice for techniques they use that you haven't learned yet. So for now just make drops negative, rises positive.

 

[moenste]

Make a drawing of the circuit. Draw in the current direction. For every resistor that the current passes through place a small + sign where it enters and a - sign where it exits the resistor. That's it. When you do your walk along a path of the circuit just account for the change in potentials using those signs. + to - is a drop so it's negative. - to + is a rise so that's positive.

Some people sum the drops and treat them as positive, while rises are accounted as negative. There's aesthetic reasons for this having to do with making the math look nice for techniques they use that you haven't learned yet. So for now just make drops negative, rises positive.

Question 1: how do we determine the current direction? At AF I looked at the 10 V and took the direction from the + side.

Question 2: voltmeter is like a resistor? Where current enters is a plus and where it goes out it's a minus?

ABEF: - V - (1 * 3) + 10 = 0. V = 7 V

BEDL: - V - 4 - (1 * 3) = 0. V = - 7 V. On LD both are minus since we are moving clockwise while the current is anti-clockwise as on the graph, therefore the signs are in the other direction.
http://tinypic.com/r/vnfmkg/9

 

[gneill]

Question 1: how do we determine the current direction? At AF I looked at the 10 V and took the direction from the + side.

If you can determine the current direction just by noting or even estimating the net EMF driving it, that's fine. But sometimes a circuit is complex and you can't determine ahead of time which way the current is going to flow in any given loop or along a given path if there are separate currents. That's fine! Just pick a direction and pencil it in. From then on just be consistent and use the same assumed direction when writing your equations. The math will sort it out for you! When you solve the equation for the current and the result turns out to be negative, then it just means that your initial guess was not right: the current actually flows in the opposite direction from your guess. The magnitude of the current you find will still be correct, and you can still plug the negative value into any equations you wrote and get correct results.

Question 2: voltmeter is like a resistor? Where current enters is a plus and where it goes out it's a minus?

Yes. Although good voltmeters generally have very high resistance and pass negligible current. Unless you are specifically analyzing a voltmeter or how it affects a particular circuit, their current is taken to be zero. Ideal voltmeters shown on a circuit to show you where to "read" a potential difference are taken to have infinite resistance and so pass no current at all (effectively they are an open circuit).

ABEF: - V - (1 * 3) + 10 = 0. V = 7 V

BEDL: - V - 4 - (1 * 3) = 0. V = - 7 V. On LD both are minus since we are moving clockwise while the current is anti-clockwise as on the graph, therefore the signs are in the other direction.
http://tinypic.com/r/vnfmkg/9

In your diagram the 3 A you've shown for the voltmeter branch is not correct. It should be zero amps. If you do KCL (Kirchhoff's Current Law) at node B then you you would see that its sum would not be zero if you had two paths of 3 A leaving and only one entering.

 

[gneill]

BEDL: - V - 4 - (1 * 3) = 0. V = - 7 V. On LD both are minus since we are moving clockwise while the current is anti-clockwise as on the graph, therefore the signs are in the other direction.

You're going clockwise around the loop, so when you pass through the 4.0 V source you're going from - to + through it. That makes it a potential rise. You're also moving against the current when you go through the resistor, so again going from - to + and that's another rise. Both those terms should be positive.

 

[moenste]

In your diagram the 3 A you've shown for the voltmeter branch is not correct. It should be zero amps. If you do KCL (Kirchhoff's Current Law) at node B then you you would see that its sum would not be zero if you had two paths of 3 A leaving and only one entering.

Ah yes, since it's a voltmeter there's no current. And since we got V = -7 V then the current is going in the opposite direction to the one we put on the graph. But we can still say that the voltmeter says V = 7 V.

+ to - is a drop so it's negative. - to + is a rise so that's positive.

This part was particularly helpful, I didn't get it before what is a drop and what is a raise.

---

I think this should be it concerning this question : ). Thank you!

 

[moenste]

You're going clockwise around the loop, so when you pass through the 4.0 V source you're going from - to + through it. That makes it a potential rise. You're also moving against the current when you go through the resistor, so again going from - to + and that's another rise. Both those terms should be positive.

Yes, but we are moving against the current. Therefore the signs should be opposite. - V - 4 (raise but opposite to the current so negative) - (3 * 1 raise but opposite to the current so negative) = 0.

 

[gneill]

Ah yes, since it's a voltmeter there's no current. And since we got V = -7 V then the current is going in the opposite direction to the one we put on the graph. But we can still say that the voltmeter says V = 7 V.

Actually, the current shown on the diagram happens to be correct. It's your equation that didn't obey the rules (was wrong)

Yes, but we are moving against the current. Therefore the signs should be opposite. - V - 4 (raise but opposite to the current so negative) - (3 * 1 raise but opposite to the current so negative) = 0.

Voltage sources are completely immune to current. They keep their potential difference as-is no matter what current, or which direction the current is flowing. So always write down their rise or fall in potential according to the +/- on the symbol as you traverse them. In this case you go from the negative to the positive pole of the battery. That's a potential rise. It would only be a drop if you walked through it in the other direction.

You are passing through the resistor also from the - to the + so again that's a rise and positive.

Remember, you put the +'s and -'s on the diagram after you picked the current direction. They are then fixed. They reflect the current's effects. After that, how you sum them only depends on your direction of travel from that point. You don't have to revisit the current direction for every component, just look at the +/- marks.

 

[moenste]

Actually, the current shown on the diagram happens to be correct. It's your equation that didn't obey the rules (was wrong) Voltage sources are completely immune to current. They keep their potential difference as-is no matter what current, or which direction the current is flowing. So always write down their rise or fall in potential according to the +/- on the symbol as you traverse them. In this case you go from the negative to the positive pole of the battery. That's a potential rise. It would only be a drop if you walked through it in the other direction.

You are passing through the resistor also from the - to the + so again that's a rise and positive.

Remember, you put the +'s and -'s on the diagram after you picked the current direction. They are then fixed. They reflect the current's effects. After that, how you sum them only depends on your direction of travel from that point. You don't have to revisit the current direction for every component, just look at the +/- marks.

I have a similar question in my book, therefore I did just like it is here ("traversing FEDCF"):

And I did the same thing...

 

[gneill]

Ah. They're using the convention that drops are summed as positive values and rises and negative ones. I mentioned this possibility in post #16 (last paragraph).

If you want to stick with that convention then just reverse the order of the +/-'s on the resistors *and* the voltage sources. All drops are positive, all rises negative.

Alternatively, leave the +/-'s alone, but assign the corresponding term in the equation the sign of the first + or - that you encounter when you walk over the component. Thus if the component was labelled + ... -, and you were traversing it from left to right, you'd hit the "+" first and would make that term positive.

Personally I find that convention a bit odd, equating a drop in potential with a positive value, but it's mathematically sound so long as its held consistently throughout the equations that are written. Again, the math takes care of everything so long as you're consistent in applying the convention.

So your equation for loop BEDCB becomes:

$V - 4 - 3 \cdot 1 = 0$
$V = 4 + 3~~~~~$ {Notice how the terms change sign when you move them over the "="}
$V = 7$ (Volts)

 

[moenste]

So your equation for loop BEDCB becomes:

And if we stick to your method it should be: - V + 4 + (3 * 1) = 0 → V = 7 V. Right? (Drop + -, then two rises - + and - +)

P. S. Sorry for so many questions concerning electricity and circuits. The book has very limited explanatory material and mostly has questions which have many nuances that aren't dealt in the book or explained properly. And since I study on my own and some topics have different approaches (like the Kirchhoff's rules here) I have to ask lots of questions which might sound dumb : (. And thanks a lot for the patience!

 

[gneill]

And if we stick to your method it should be: - V + 4 + (3 * 1) = 0 → V = 7 V. Right? (Drop + -, then two rises - + and - +)

Right.

P. S. Sorry for so many questions concerning electricity and circuits. The book has very limited explanatory material and mostly has questions which have many nuances that aren't dealt in the book or explained properly. And since I study on my own and some topics have different approaches (like the Kirchhoff's rules here) I have to ask lots of questions which might sound dumb : (. And thanks a lot for the patience!

No worries. I and the other helpers here are happy to help students who how effort and want to learn.