# Find initial current and resistance in a circuit

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1. Oct 2, 2016

### moenste

1. The problem statement, all variables and given/known data
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)?

2. Oct 2, 2016

### 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 ?

3. Oct 2, 2016

### Twigg

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.

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.

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

4. Oct 3, 2016

### moenste

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

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

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, RX = 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.

5. Oct 3, 2016

### BvU

Would it be easier for you if you represented the non-ideal voltage source as an ideal voltage source with an internal resistance in series ?

6. Oct 3, 2016

### BvU

That way it's easier to answer

7. Oct 3, 2016

Like this?

8. Oct 3, 2016

### 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 )

9. Oct 3, 2016

### moenste

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.

10. Oct 3, 2016

### 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 Rs, the internal resistance.

11. Oct 3, 2016

### moenste

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

Update:

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

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".

Last edited: Oct 3, 2016
12. Oct 3, 2016

### 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 Rs

13. Oct 3, 2016

### moenste

Hm, but this graph:

Clearly states that a non-ideal voltage source has a VS and a RS of Vd. And an ideal one doesn't have RS. 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.

Last edited: Oct 3, 2016
14. Oct 3, 2016

### BvU

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

15. Oct 3, 2016

### moenste

We have VVoltmeter = 12 V and VSource + VResistor 0.5 Ω = 12 V (because it's parallel).

I RSource + 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 VSource + VResistor = 12 V stays the same) while it's the VSource that decreases by 2 V? So something like

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

Last edited: Oct 3, 2016
16. Oct 3, 2016

### BvU

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

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

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 Rs

17. Oct 3, 2016

### moenste

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

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?

Last edited: Oct 3, 2016
18. Oct 3, 2016

### BvU

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.

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\$).

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.

19. Oct 3, 2016

### moenste

There is the same potential difference across each line.

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

20. Oct 3, 2016

### moenste

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

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

Last edited: Oct 3, 2016
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