Find initial current and resistance in a circuit

[moenste]

Homework Statement

A battery of internal resistance 0.50 Ω is connected (as shown below) through a switch S to a resistor X, which is initially at 0 °C. When S is closed, the voltmeter reading falls immediated from 12.0 V to 10.0 V. The reading then rises gradually to a steady value of 10.5 V.

(a) Explain these observations.
(b) Calculate: (i) the initial current when S is closed; (ii) the initial resistance of X; (iii) the resistance of X when he steady state has been reached.

Answers: (b) (i) 4.0 A, (ii) 2.5 Ω, (iii) 3.5 Ω.

2. The attempt at a solution
(a) Maybe it is the temperature that gives an increase over time to 10.5 V? The initial temperature increases from zero to some unknown value.

(b) (i) I tried to find the current via I = V / R = 12 / 0.5 = 24 A, but the answer is wrong.

(b) (ii) E = I (r + R) → 12 = 4 (from the answer) * (0.5 + R) → R = 2.5 Ω.

(b) (iii) I guess I'll need a new current to correctly calculate this part. 10.5 = 4 (0.5 + R) → R = 2.125 Ω, which is wrong.

So it's mostly about (a) and (b) (i). Is it due to the temperature in (a) and how to find current in (b) (i)?

 

[BvU]

Hi,

Starting with (b) is sensible thing to do. Your 24 A disregards the resistance X. You are supposed to compare the 12 V, 0 A case with the 10 V case when current flows. Where does the 2 V voltage drop occur ?

 

[Twigg]

(a) Maybe it is the temperature that gives an increase over time to 10.5 V? The initial temperature increases from zero to some unknown value.

Good thinking! You may want to add that for an ordinary conducting material, the resistance is expected to increase with temperature. How can you reconcile this with the observation that the voltage drifted from 10V to 10.5V? Also you may want to explain why the temperature of X goes up.

(b) (i) I tried to find the current via I = V / R = 12 / 0.5 = 24 A, but the answer is wrong.

The 0.50$\Omega$ resistor and the load X are connected in series when the switch is closed. Is there really 12V across the 0.50$\Omega$ resistor? Keep in mind that the battery puts out a maximum of 12V when it is not supplying a current and that the terminal voltage (which is measured after the 0.50$\Omega$ resistor, hint hint) drops to 10V when it is supplying the unknown current.

(b) (iii) I guess I'll need a new current to correctly calculate this part. 10.5 = 4 (0.5 + R) → R = 2.125 Ω, which is wrong.

Does this current go through the 0.50$\Omega$ resistor? Remember the voltage is measured after!

 

[moenste]

How can you reconcile this with the observation that the voltage drifted from 10V to 10.5V?

Hm, I think it should be like I = 20 A, R_Initial = 15 Ω and R_Final = 16 Ω. So V_Initial = I R = 20 * 15 = 300 V and V_Final = 20 * 16 = 320 V. So with the increase in resistance the voltage also increases. [Numbers are solely for example purposes.]

Also you may want to explain why the temperature of X goes up.

Well, maybe due to the current that starts flowing though the resistor and it heats it up?

The 0.50Ω\Omega resistor and the load X are connected in series when the switch is closed. Is there really 12V across the 0.50Ω\Omega resistor? Keep in mind that the battery puts out a maximum of 12V when it is not supplying a current and that the terminal voltage (which is measured after the 0.50Ω\Omega resistor, hint hint) drops to 10V when it is supplying the unknown current.

Starting with (b) is sensible thing to do. Your 24 A disregards the resistance X. You are supposed to compare the 12 V, 0 A case with the 10 V case when current flows. Where does the 2 V voltage drop occur ?

I don't quite understand how we can find current or the resistance of X.

So we have V = I R or E = I (r + R). In the first case: V =12 V, I = 0 A, r = 0.5 Ω, R_X = X so we have 12 = 0 (0.5 + X). No idea how to apply this equation. The second one we have: 10 = I (0.5 + X). Two unknowns.

 

[BvU]

Would it be easier for you if you represented the http://www.designcabana.com/knowledge/electrical/basics/sources/ source as an ideal voltage source with an internal resistance in series ?

 

[BvU]

That way it's easier to answer

Where does the 2 V voltage drop occur ?

 

[moenste]

Would it be easier for you if you represented the http://www.designcabana.com/knowledge/electrical/basics/sources/ source as an ideal voltage source with an internal resistance in series ?

Like this?

 

[BvU]

Yes. Better draw the 12 V as an ideal voltage source now (see link) and the 0.5 $\Omega$ closer to that voltage source. Note that the voltmeter measures the voltage over X, not over the X + 0.5 $\Omega$ (because that is still ... V )

 

[moenste]

Yes. Better draw the 12 V as an ideal voltage source now (see link) and the 0.5 $\Omega$ closer to that voltage source. Note that the voltmeter measures the voltage over X, not over the X + 0.5 $\Omega$ (because that is still ... V )

Isn't it the ideal voltage source? The internal resistance is included in the battery.

I thought the voltmeter measured voltage over 0.5 Ω and therefore it has 12 or 10 or 10.5 V since they are in parallel.

 

[BvU]

Read the link (at "To show non ideality of a voltage source, they add a series resistor to its output, ...") !

The voltmeter measures the voltage at the clamps of the non-ideal voltage source (= 12 V battery) so: what comes out of the dashed box. Namely the 12 V minus the voltage drop over R_s, the internal resistance.

 

[moenste]

Read the link (at "To show non ideality of a voltage source, they add a series resistor to its output, ...") !

The voltmeter measures the voltage at the clamps of the non-ideal voltage source (= 12 V battery) so: what comes out of the dashed box. Namely the 12 V minus the voltage drop over R_s, the internal resistance.

I read it all. You say we have a non-ideal case. And you say we need to make it an ideal case.

Update:

Would it be easier for you if you represented the http://www.designcabana.com/knowledge/electrical/basics/sources/ source as an ideal voltage source

So we have a non-ideal source that needs to be presented as an ideal voltage source.

Better draw the 12 V as an ideal voltage source now

And again you state that we need to draw as an ideal voltage source.

As I understood, a non-ideal has PD + resistance, while an ideal has just PD:

And this is why I combined to whole thing into one to make it an "ideal source".

 

[BvU]

No. We just replace the battery (which is non-ideal voltage source with internal resistance -- as given in your exercise)

by an ideal voltage source with an internal resistance R_s

 

[moenste]

No. We just replace the battery (which is non-ideal voltage source with internal resistance -- as given in your exercise)

View attachment 106872

by an ideal voltage source with an internal resistance R_s

View attachment 106873

Hm, but this graph:

Clearly states that a non-ideal voltage source has a V_S and a R_S of V_d. And an ideal one doesn't have R_S. At least I understand this graph as everything that is inside the frame is the non-ideal voltage source.

Update:

Maybe like this?

But this is how I imagined the graph from the beginning. Though I don't see how it helps. Maybe Because the voltmeter shows 12 V in the beginning, we have the battery V + 0.5 Ohm resistor = 12 V? But that doesn't help much, since we already have 12 = 0 (0.5 + X) and 10 = I (0.5 + X) either zero or two unknowns.

 

[BvU]

What is the voltage drop over the $0.5\ \Omega$ when there is no current flowing ?

 

[moenste]

What is the voltage drop over the $0.5\ \Omega$ when there is no current flowing ?

We have V_Voltmeter = 12 V and V_Source + V_{Resistor 0.5 Ω} = 12 V (because it's parallel).

I R_Source + I * 0.5 = 12.

Maybe when the voltage decreases to 10 V it's not the 0.5 Ω resistor that decreases (it's unknown proportion in the V_Source + V_Resistor = 12 V stays the same) while it's the V_Source that decreases by 2 V? So something like

(I R_Source - 2) + I * 0.5 = 10.

 

[BvU]

We have V_Voltmeter = 12 V and V_Source + V_{Resistor 0.5 Ω} = 12 V (because it's parallel).

V_Source = 12 V so V_{Resistor 0.5 Ω} must be 0 V !

I R_Source + I * 0.5 = 12.

No. The $0.5\ \Omega\ \$ is the R_s

When there is no current flowing, the meter reads 12 V, just as much as the voltage source delivers: there is zero voltage drop over R_s

 

[moenste]

VSource = 12 V so VResistor 0.5 Ω must be 0 V !

Why? How I see it: we have a parallel voltmeter with V = 12 V. The voltmeter is parallel to the V_Source AND the resistor 0.5 Ohm. Therefore the V_Source and the 0.5 Ohm resistor have 12 V as their voltage due to the parallel circuit rule.

VSource = 12 V so VResistor 0.5 Ω must be 0 V !

No. The 0.5 Ω 0.5\ \Omega\ \ is the Rs

V = 12 V
R = 0.5 Ohm
I = V / R = 12 / 0.5 = 24 A. Like the one I got in the very first post.

If we do 12 - 10 = 2 V. Then I = V / R = 2 / 0.5 = 4 A. Is this correct?

 

[BvU]

Why? How I see it: we have a parallel voltmeter with V = 12 V. The voltmeter is parallel to the V_Source AND the resistor 0.5 Ohm. Therefore the V_Source and the 0.5 Ohm resistor have 12 V as their voltage due to the parallel circuit rule.

I'm not familiar with the terminology 'parallel circuit rule', sorry. Both ends of the $0.5\ \Omega$ resistor are at +12 V. There is no voltage drop over the resistor, consistent with no current flowing.

V = 12 V
R = 0.5 Ohm
I = V / R = 12 / 0.5 = 24 V. Like the one I got in the very first post.

That would be the short-circuit current. Not 24 V, but 24 A. What you would get if the clamps of the battery were connected directly with one another ($X=0\ \Omega\$).

If we do 12 - 10 = 2 V. Then I = V / R = 2 / 0.5 = 4 A. Is this correct?

This is the next step: S is closed and the voltmeter reads 10 V.

So the left end of the $0.5\ \Omega$ sees 12 V and the right end 10 V. The voltage drop is 2 V indeed.

 

[moenste]

I'm not familiar with the terminology 'parallel circuit rule', sorry.

There is the same potential difference across each line.

Both ends of the 0.5 Ω0.5\ \Omega resistor are at +12 V. There is no voltage drop over the resistor, consistent with no current flowing.

The resistor is in series with the power source. Therefore it has V_Source + V_resistor = 12 V, because there is the same potential difference across each line.

 

[moenste]

Calculate: (i) the initial current when S is closed

We have r = 0.5 Ohm and V_voltmeter = 12 V.

Since the voltmeter is connected to the battery we can ignore the R_X resistor. Then because the voltmeter is connected to the batter in parallel, we know that the V_battery = also 12 V. So we can calculate the initial current I = V_battery / R_battery = 12 / 0.5 = 24 A.

 

[BvU]

There is no current. Where would it go ?

And 12 V / 0.5 Ohm = 24 A, not 24 V !

 

[moenste]

There is no current. Where would it go ?

And 12 V / 0.5 Ohm = 24 A, not 24 V !

So the answer is zero?

 

[BvU]

What question would that be 'the' answer to ? There is no current when S is open. That's how the voltmeter tells you what V_s is.

When S is closed, you do have a current, and the voltmeter reads less than V_s, namely V_o = 10 V in your exercise. What does that tell you about V_d ?

 

[moenste]

What question would that be 'the' answer to ? There is no current when S is open. That's how the voltmeter tells you what V_s is.

When S is closed, you do have a current, and the voltmeter reads less than V_s, namely V_o = 10 V in your exercise. What does that tell you about V_d ?

(b) Calculate: (i) the initial current when S is closed;

When S is closed the initial voltage of the voltmeter is 12 V and internal resistance is 0.50 Ω.

1. The voltmeter that has 12 V is parallel to the battery.
2. The battery has a voltage of 12 V as well. Because in parallel circuits voltage is equal across the lines.
3. I = V / R = 12 / 0.5 = 24 A.

 

[BvU]

When S is closed the initial voltage of the voltmeter is 12 V and internal resistance is 0.50 Ω.

Ah, I see the source of misunderstanding. The exercise composer is more to blame than you.

When S is closed, the voltmeter reading falls immediately from 12.0 V to 10.0 V.

By that he means that the voltmeter reads 10 V 'initially' for the answer to question (b).

(Relatively long) after that, the resistor heats up and the meter gradually goes to 10.5 V

1. The voltmeter that has 12 V is parallel to the battery.

It is, but it reads 10 V. And the battery consists of a series circuit of a 12 V voltage source and an internal resistance.

2. The battery has a voltage of 10 V as well

Yes (10 V) which consists of 12 Volt V_s minus a voltage drop over the internal resistance.

3. I = V / R = 12 / 0.5 = 24 A.

No. the 12 V is distributed over the series circuit consisting of R_s and X.
And we know the voltage over X (namely 10 V, given)

To relieve me of a worrying concern:
Is the relevant section of the link (at "To show non ideality of a voltage source, they add a series http://www.designcabana.com/knowledge/electrical/basics/resistors to its output, ...") completely clear to you ?

 

[moenste]

By that he means that the voltmeter reads 10 V 'initially' for the answer to question (b).

(Relatively long) after that, the resistor heats up and the meter gradually goes to 10.5 V

Hm, what you suggest I should read the problem this way: we have a battery of 12 V, after the circuit is turned on the voltmeter shows 10 V? So it's only the voltmeter that changes and the power source remains the same? And then we just calculate E = Ir + V → 12 = 0.5 I + 10 → 2 = 0.5 I → I = 4 A?

 

[moenste]

To relieve me of a worrying concern:
Is the relevant section of the link (at "To show non ideality of a voltage source, they add a series http://www.designcabana.com/knowledge/electrical/basics/resistors to its output, ...") completely clear to you ?

I was looking for the phrase, but the link was a different one from that on the first page.

OK we have:

To show non ideality of a voltage source, they add a series http://www.designcabana.com/knowledge/electrical/basics/resistors to its output, therefore with more current drawn out of the source, more voltage drops across the source resistor and decreases the output level of the source. Below a non-ideal voltage source is connected to a resistor.

An IDEAL voltage source would be just V_S. Just the circle with + and -.

In our case we have a NON-ideal voltage source and therefore everything INSIDE the dotted / striped area is regarded as the non-ideal voltage source: V_S + V_d = V_O.

For this problem:
To show non ideality of a voltage source, they add a series resistor (0.5 Ohm) to its output, therefore with more current drawn out of the source (12 V), more voltage drops across the source resistor and decreases the output level of the source (10 V).

This is how I understand this.

 

[BvU]

Sounds good to me -- provided you realize the source resistor is the series resistor is the internal resistor is the 0.5 Ω
And the 10 V is the output of the combination source & internal resistor

So how much current is needed to make this output level 10 V ?

 

[moenste]

Sounds good to me -- provided you realize the source resistor is the series resistor is the internal resistor is the 0.5 Ω
And the 10 V is the output of the combination source & internal resistor

So how much current is needed to make this output level 10 V ?

Hm, what you suggest I should read the problem this way: we have a battery of 12 V, after the circuit is turned on the voltmeter shows 10 V? So it's only the voltmeter that changes and the power source remains the same? And then we just calculate E = Ir + V → 12 = 0.5 I + 10 → 2 = 0.5 I → I = 4 A?

I think this should be it for (b) (i)?

(b) (ii) is clear to me.

However, how to find (b) (iii)? As I understand we'll need a new current? 10.5 = I (X + 0.5). Or like 12-10.5 = 0.4 X so X = 3.75 Ohm.

 

[BvU]

That's it for b(i).
b(ii) gives what resistance ?

In b(iii) you go off the rails, I'm afraid. The 10.5 V is over resistance X, NOT over the series circuit of X + 0.5 Ω !

Or like 12-10.5 = 0.4 X so X = 3.75 Ohm

Where on Earth would that 0.4 come from ?

 

[moenste]

b(ii) gives what resistance ?

(b) (ii) E = I (r + R) → 12 = 4 (from the answer) * (0.5 + R) → R = 2.5 Ω.

In b(iii) you go off the rails, I'm afraid. The 10.5 V is over resistance X, NOT over the series circuit of X + 0.5 Ω !

So the resistor X has a voltage of 10.5 V?

 

[BvU]

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current throught the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

 

[moenste]

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current throught the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

We have a voltage battery 12 V. We have a resistor -2 V. We have a voltmeter 10 V. We have a resistor X 0.5 V and current I = 0.4 A. So R = V / I = 0.5 / 0.4 = 1.25 Ohm.

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current throught the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

12 - 10.5 = 0.5 I
I = 3 A new current in the parallel circuit?

 

[BvU]

We have a voltage battery 12 V. We have a resistor -2 V. We have a voltmeter 10 V. We have a resistor X 0.5 V and current I = 0.4 A. So R = V / I = 0.5 / 0.4 = 1.25 Ohm.

You confuse me. The X resistor sees 10 V. That's what the meter indicates. How can it possibly be 0.5 V !? Where does the 0.4 A come from ?

 

[moenste]

You confuse me. The X resistor sees 10 V. That's what the meter indicates. How can it possibly be 0.5 V !? Where does the 0.4 A come from ?

So the V_X = 10 V. And internal resistance of the battery is 0.5 Ohm. How do we find the R_X?

(i) 12 - 10 = 0.5 I → I = 4 A.

(ii) 12 = 4 (0.5 + R) → R = 2.5 Ω

(iii) 10.5 = 4 (0.5 + R) → R = 2.1 Ω

or 12 - 10.5 = 0.5 I → I = 3 A
10.5 = 3 (0.5 + R) → R = 3 Ω

or 12 - 10.5 = 0.5 I → I = 3 A
12 = 3 (0.5 + R) → R = 3.5 Ω