# Help with AC and DC

1. Sep 28, 2006

### tdaffin

Hi, it's been a long time since I did any physics, and electricity was never my strong point.

I've been revising the basics from an experimental standpoint, and I've run into something that I don't understand. I hope someone here will be able to help.

I have a 75W 120V light bulb, which I expect (by calculation) to have a resistance of 192ohms, and hence pass a current of 0.625A when hooked up to a 120V AC supply. I've done that and tested the configuration and my multimeter cofirms that current is being passed.

Then, I made the assumption that if I hooked the same light bulb up to a 12V DC car battery, it would have the same resistance, and hence pass a current of 0.0625A.

When I hooked it up in that configuration and measured the DC current, I got a measurement of 0.18A, which I think implies a resistance of 66ohms.

What is going on here? Does the lightbulb have a different resistance to DC than AC?

One thing to note here is that in the 12V DC circuit, the bulb did not light up. Perhaps then this is just shows the temperature dependance of resistance showing up?

2. Sep 28, 2006

### tdaffin

One more thing... I went back to the garage after posting this and measured the resistance of the bulb and got 13ohms. So, now we have 192ohms(calculated) @120v (illuminated), 66ohms @12v (no illumination) and 13ohms @ 0v (room temperature). I guess it would help if I could get a temperature on the buld in all 3 cases. Is there a particular shape to the graph that I should expect? Linear? Logarithmic? Other?

3. Sep 28, 2006

### Rozenwyn

The graph is logarithmic but around room temperature it shouldn't vary much. And the calculated resistance for the lightbulb should be the base resistance (or whatever it is called) of it, i.e. the resistance at room temperature.
Let's see:
$$P = VI \ \longrightarrow I = 0.625A$$ and $$V=IR \ \longrightarrow \frac{120}{0.625} = 192 \Omega$$. You got this.
When you vary the P.D., resistance doesn't change but current does. You need to realize that current passing through this bulb doesn't always equal 0.625A.
So in case of a 12V DC battery $$I=\frac{12}{192}=0.0625A$$, this implies that less power is produced which is insufficient to light the bulb.

I can't think of a reason why your multimeter would measure a resistance of 13 Ohms.

4. Sep 29, 2006

### Averagesupernova

All three are correct. The filament resistance was most certainly 192 ohms when hooked up to 120 volts. It was also 66.666 ohms when hooked to a car battery. And when at room temp I'm sure it was 13 ohms. The different temperatures of the filament in all three cases causes the actual resistance of the filament to vary.
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Rosenwyn, I have no idea what you are talking about. I can tell you that Ohms law ALWAYS works.
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One last thing, it has nothing to do with AC vs. DC.

5. Sep 29, 2006

### tdaffin

Thank you both for your help!

I guess I need to devise a method to measure the filament temperature in all 3 circumstances and then plot that vs. the resistance.

If it comes out logarithmic (with 3 data points, that won't be so convincing -- just not linear), I'll rest my case for now.

6. Sep 29, 2006

### Staff: Mentor

:rofl: :rofl: :rofl: I'm glad I'm not the only one who looked at that and went, huh?

7. Sep 29, 2006

### Staff: Mentor

Can't you use the ohmmeter when the power is switched on to verify the calculations?

8. Sep 29, 2006

### Staff: Mentor

You can, but keep in mind that for the current measurements, the meter inserts a series resistance in the circuit. If you know what that resistance is, then you can make fairly accurate current and voltage measurements.

9. Sep 29, 2006

### waht

The resistance of the filament changes from heat when it is glowing.

So when you measure current at 120 V, the light bulb is glowing already so it has a different resistance, than when it is at 12 V. Because 12 V is not enough to heat the filament.

10. Sep 29, 2006

### Averagesupernova

I hope you misunderstood russ' question. An ohmeter is never EVER used when the power is on. It provides its own power. A sure way to damage a cheap analog ohmeter is to probe across a voltage with it.

11. Sep 29, 2006

### tdaffin

Well, thanks again everyone... I was going to go home and put my ohmmeter across the live bulb!
Thanks to berkeman for informing me that the ammeter will have a series resistance -- I have two multimeters so I can use one to measure the resistance of the other when it's an ammeter.
Thanks also to Averagesupernova for saving my ohmmeter.
So, I guess I'm back to trying to measure the filament temperature and seeing if the R vs. T graph is not linear. Ideally I'd like to plot a curve through the points and then set things up so that the T will be another value, measure the current flow in that circuit and see if the calculated reistance falls on the curve again.

12. Sep 29, 2006

### tdaffin

Yes, thanks waht, this also fits with the bulb having a lower resistance again when no voltage is across it at all (measuring it using the ohmmeter).
I guess that even though it's not hot enough to produce light with 12v across it that it has heated up above room temperature -- enough to raise the reisistance from 13ohms to 66.

13. Sep 29, 2006

### Staff: Mentor

Not so much a misunderstanding, just loose use of terminology by Russ and me (well, at least me). Terms like meter, ohmeter, voltmeter, DVM, etc., get used synonomously often, unless there is a very specific context. Pretty much all DVMs have V, I, R functionality, and often more. So I was making the assumption that Russ was just referring to a general DVM, configured in current measuring mode. Truth be told, I didn't even consciously register that I was making the translation. But you're right supernova, in resistance measurement mode, it might not be a good idea to connect it in series with the bulb....

14. Sep 29, 2006

### tdaffin

Ok, thanks for clearing that up. I understand now that it's a bad idea to hook the DVM up in series when it's in resistance mode.

Is it possible to use the DVM in parallel across the component in resistance mode when the circuit is powered up? One thing that springs to mind for me is that I wouldn't know if I was measuring the resistance of that component, or of the rest of the circuit...

15. Sep 29, 2006

### NoTime

No!
The resistance/ohmeter function supplies a small amount of power to the component and reads the drop acrost a calibrated resistance inside the unit.

Applying any external voltage source will most likely damage the meter.
At the very least you will get an invalid measurement.

16. Sep 30, 2006

### tdaffin

Ok, thanks. I'll consider that the last answer then. So, I shall not be using my Multi-meter to measure resistances in any energised circuits.