Why does this happen?

Just before we begin, I am fairly new to electronics.

I have an input of 8.2V from a battery (checked using multimeter). When I connect three resistors in series (each of 180Ω) with this voltage source, my multimeter reads 7.1V across the three resistors. What is happening to the other 1.1V?
I thought that resistors were supposed to equally share the voltage for the source.



Also on a side note, I had my step down transformer hooked up to a full wave bridge rectifier rated at 1.5A 500V. I put three resistors in series connected to my source. V=19 Req=~545 therefore I=34mA
Now everything was running smoothly until I tried to measure the current across the resistors using a multimeter. Like two seconds later (while my multimeter was still reading the current) my full wave bridge rectifier started smoking.
Why did this happen? Was it because I tried to measure the current and that some how overloaded the current? Or is my rectifier not fit to do the job? Sorta scared me, don't want to use my transformer anymore in lieu of a flare up/something blowing up in my face.

Thank you for your help/time.

The battery itself has an internal resistance (not an actual resistor but an unavoidable way that it works).

To measure current you have to break the circuit and put the multimeter probes in the gap. If you try to measure current the same way you measure voltage then the multimeter has a very low resistance and so lots of current will flow through the multimeter. Hopefully the fuse in the multimeter blows quickly enough to prevent damage, but not always. Of course if you moved the probe to the unfused terminal of the multimeter then bad things happen.

1) The battery has some internal resistance which drops some voltage at its terminals when under load. The voltage you are missing is dropped across this internal resistance. As a battery is depleted, its internal resistance goes up. Your meter has a very high resistance so measuring the battery voltage with no other load drops a negligible voltage across its internal resistance.

2) Measuring amps requires that the meter be in series (not parallel across the resistors). Ammeters have a very low resistance so as not to affect the circuit under test. Setting the meter to amps and 'measuring' across the battery puts a very small resistance (the meter) in parallel with a much greater resistance (the resistors). Most of the current will flow through the meter. It doesn't like that. Most ammeters are fuzed to protect against just this sort of thing.

Edit:
Dang, Mark, you type pretty fast..

Thanks.
But when I hook up a 1.5V battery to the resistors the voltage does read 1.5V across the three resistors. Does this mean there is almost 0 internal resistance in the 1.5V battery, or will the voltage of the battery stop being equal to the voltage across the resistors when the battery starts to lose voltage?
And from what I understand, my full wave bridge rectifier only broke because I measured the current incorrectly, not because it is not fit to do the job, correct?

Once again, thank you. I would have kept on measuring current the same way if I hadn't been told this.

But when I hook up a 1.5V battery to the resistors the voltage does read 1.5V across the three resistors. Does this mean there is almost 0 internal resistance in the 1.5V battery, or will the voltage of the battery stop being equal to the voltage across the resistors when the battery starts to lose voltage?

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It means that the internal R of the battery is significantly less than the R's and it is able to provide sufficient current to keep the terminal voltage more or less constant. But all batteries have internal resistance, a bigger battery has less than a small one. That's why you can't start your car with a 12 volt garage door opener battery.

And from what I understand, my full wave bridge rectifier only broke because I measured the current incorrectly, not because it is not fit to do the job, correct?

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Correct. You essentially shorted the output of the bridge with the meter. Consider fuzing that as well.

EDIT: this assumes that the bridge was sufficient in the first place..

Several things are going on.
1. A 1.5V AA will have 1/6 the resistance of 6 of them in series (9V).
2. A 9V PP3 battery will generally have higher resistance than 6 AAs in series, because the individual cells (there are 6 of them) of the PP3 are smaller.
3. If your 8.1V battery is alkaline, it is near the end of its life. Internal resistance goes up as batteries get used up.
4. 1.5V pushes 1/6 as much current through a given resistance as does 9V, so the voltage drop (I*R) of the internal resistance will be less,even if the resistances are equal.

JohnInTX said:

EDIT: this assumes that the bridge was sufficient in the first place..

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Just to double check with you because I've only been doing this for < 2weeks. Rectifier is rated at 1.5A, 100V. Source is ~18.9V DC. Req = ~545Ω. So I = 18.9/545 = 0.0346A. Nothing exceeds rectifier ratings. Correct?

last question,
If I were to put resistors in series with an LED, the entire series would have the current of the LED?

and

if I had resistors in series with two LEDs with two different currents, what would the current of the series be?

Then of course there is the accuracy of your meter to consider.

Just to double check with you because I've only been doing this for < 2weeks. Rectifier is rated at 1.5A, 100V. Source is ~18.9V DC. Req = ~545Ω. So I = 18.9/545 = 0.0346A. Nothing exceeds rectifier ratings. Correct?

Click to expand...

It should be OK. For the record, I always over spec these. For a couple of cents more you can get to 400V.

If I were to put resistors in series with an LED, the entire series would have the current of the LED?

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Yup. Current in a series connection is the same through every device. Voltages across each device will vary.

if I had resistors in series with two LEDs with two different currents, what would the current of the series be?

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If you are asking if each LED/R was in series then these were in parallel, the current through each parallel branch would depend on the LED Vforward and the resistor. If you are placing Rs and LEDs in complete series, the previous note would apply.

Look through the ebooks / tutorials on this site. They are very well written and will answer most of your questions.

Gotta go. thermometer beeped and time to take the lamb out of the smoker. Yum!

davidGG said:

Just to double check with you because I've only been doing this for < 2weeks. Rectifier is rated at 1.5A, 100V. Source is ~18.9V DC. Req = ~545Ω. So I = 18.9/545 = 0.0346A. Nothing exceeds rectifier ratings. Correct?

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The specs are probably talking about two different operating conditions, namely that it can withstand 100V of peak inverse voltage (PIV) before breaking down and conducting current and that, when forward biased and conducting current, it can tolerate 1.5A of current. The voltage drop is probably about 0.5V to 1V under those conditions and that spec may also be if the rectifier has a heat sink. We'd have to have the detailed specs to be sure.

If I were to put resistors in series with an LED, the entire series would have the current of the LED?

Click to expand...

Yes, that is what is meant/required for two components to be considered "in series".

if I had resistors in series with two LEDs with two different currents, what would the current of the series be?

Click to expand...

Need to describe this a bit better, keeping in mind that if two components can have different currents then, pretty much by definition, they are not in series. A simple diagram (even just thrown together in Paint) would be quite useful, here.