Replaced the battery for multimeter.
I will not crack open the battery box.
Did try to short the terminals but not apparent effect.
Call exorcists ?
I'd say call someone to do a JREF challenge application.
Paranormal Battery box
- Thread starter Jyera
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I will not crack open the battery box.
Did try to short the terminals but not apparent effect.
Call exorcists ?
I'd say call someone to do a JREF challenge application.Batteries are not dead, I have bring them out of the battery box. Don't have a bulb. Will try to see if I can find one.No, as "69" (love that handle) has already explained, a very high resistance will still measure a voltage even though it will not allow a usable current (V=RA), although it is still possible that there is another short somewhere else between the batteries, as I suggested checking for.
One other simple way that occurs to check that is to put a small 1.5v flashlight bulb across the terminals when the switch is OFF. Your 1.49v should light it up, unless it is a very high resistance short, like 20 mohms in the switch.
BTW, by now your batteries must be dead. Shall we start again with new batteries and boost it up to 6v?
Simon Bridge
Critical Thinker
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Surely you mean V=IR ? Ohm's Law? But I'll let you off since the text of that message is still goodElind said:
Jyera: I don't know why you won't take advise.
This is no way to measure the internal resistance of a battery ... you measure the open circuit voltage... fine... then you measure the voltage across a known load. Then you do the math.
You do not ever use the ohms setting on a live circuit. That is a good way of breaking the meter.
What does the voltage read when the terminals are held in the air? In your mouth?Jyera said:
eg.
My Multimeter is an Escort EDM-1155
Set to DC mV range, I get fluctuating voltage of -1.2mV to -1.6mV, and the terminals are 40cm apart in air. I'm not even touching them, they are suspended by their wires from my ruler.
If I set the AC mV range, I get 9.3mV to 12.8mV.
If I touch the terminals, the readings go wild!
Interestingly, if I stick them in my mouth and hold real still, I get 0.3mV AC stable.
Most people think a voltmeter measured voltage - but this is incorrect. What a voltmeter actualy does is compare voltages between the two terminals. It takes the voltage at the red terminal and subtracts the voltage at the right terminal. Strictly and pedantically speaking, the voltmeter measures potential difference.
Now for your batteries.
All real batteries can be modelled as an ideal voltage source in series with a resistor. The resistance of the resistor is called the "internal" resistance. The voltage of the source is sometimes called the EMF. (There are other conventions - and physicists tend to dislike the term "EMF".)
The EMF is what the voltmeter says when you stick the battery between the terminals. The internal resistance is calculated from this and what the voltage is across a known load. You use Ohm's Law and the resistor-combination rules.
If the EMF is U, the internal resistance is r, the load is R, and the voltage across the load is V, then:
U=I(r+R)
V=IR
Eliminate I from the above, and rearrange:
UR=V(r+R)
... solve for r.
......
An Ohmmeter works, in essence, by applying a voltage across the subject, and measuring the current. Then divide the voltage by that current to get the resistance. If you apply the ohmmeter across a battery - the voltage applied to the internal resistance will by the meter voltage (usually about 9V) and the EMF combined. But the meter is only using the meter voltage for the calculation (it dosn't know the EMF) so it will get the calculation wrong.
Do it the standard way - you won't mess up.
Thank you Simon.
On internal Resistance.
There was so much advise from various people. Since I did not have a "known load" thus I did not carry out what you suggested. Will be trying to get some resistor. Then i'll be able to do what you suggested.
I didn't know measuring Ohm over live circuit could be harmful to the meter.
Let me understand. An individual battery isn't a circuit. But because it a component that can provide an EMF. It will affect, disrupt and perhaps damage a Ohm-meter. An Ohm-meter by itself, will supply EMF to measure the resisitance of a passive load like a resistor.
Is that right?
About Potential Difference:
I would say, being able to measure a 0.3mV within our mouth,
do not in any way diminish the fact that we are able to measure a voltage across the passive switch. The voltage is real, isn't it ?
I tried shaking my digital multimeter and it indeed show some very small voltage. The fact that the "switch@off" without battery shows a very small voltage, based on what you mentioned, it is simply showing a potential difference across the switch terminal.
But why is there a potential difference? Is it ..
(1) The multimeter is unreliable for such a small voltage?
(2) Potential diff between different type of metal contact?
There ought to be a good reason for the potential difference.
I have tried to short the terminal of the switch to drain any possible charge, but it still register a voltage.
On internal Resistance.
There was so much advise from various people. Since I did not have a "known load" thus I did not carry out what you suggested. Will be trying to get some resistor. Then i'll be able to do what you suggested.
I didn't know measuring Ohm over live circuit could be harmful to the meter.
Let me understand. An individual battery isn't a circuit. But because it a component that can provide an EMF. It will affect, disrupt and perhaps damage a Ohm-meter. An Ohm-meter by itself, will supply EMF to measure the resisitance of a passive load like a resistor.
Is that right?
About Potential Difference:
I would say, being able to measure a 0.3mV within our mouth,
do not in any way diminish the fact that we are able to measure a voltage across the passive switch. The voltage is real, isn't it ?
I tried shaking my digital multimeter and it indeed show some very small voltage. The fact that the "switch@off" without battery shows a very small voltage, based on what you mentioned, it is simply showing a potential difference across the switch terminal.
But why is there a potential difference? Is it ..
(1) The multimeter is unreliable for such a small voltage?
(2) Potential diff between different type of metal contact?
There ought to be a good reason for the potential difference.
I have tried to short the terminal of the switch to drain any possible charge, but it still register a voltage.
bruto
Penultimate Amazing
Yes, that's about it. But the voltage is small and the circuit senstive, which is why you don't want to do it on a live circuit. Most digital meters will have fuses to protect the circuitry. Making that mistake on an analog meter will sometimes damage the meter mechanism.
I'm guessing the multimeter is simply not reliable without some slight load. I've not usually had this problem with DC meters but it's very common on AC. The meter itself has very high impedance, so it does not load the circuit by itself. I'm guessing that if you put a very small load on the output of the box, the anomaly will disappear, unless there is indeed a fault somewhere in the box where you can't see it.
Here's a thought: have you tried using the milliameter scale to read the output of the box when it's off? Find the highest range of DC milliamps and measure output with the switch off. An ammeter does load a circuit, so this should tell you in a hurry if there's any real current in there.
Ririon
Cool cat
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Anyhoo... Elind got it right.
Just to make 100 % sure: What you want to measure is the resistance of the switch with the switch in the off position. When you tried to do that, you took the batteries out (right). You didn't measure the resistance between the two wires (wrong), did you?. This obviously shows open circuit, since the circuit is supposed to go through the batteries...
I think you will find that the switch connects the red wire to the + pole of the last battery (the one on the right in the picture.) Take out the batteries and measure the resistivity between the red wire and this contact point. What you will find is that with the switch on, it will be very low. Your meter will probably show zero. No problem. With the switch off, it will show a value of about the size of the meter's internal resistance in voltage mode. Very high, but it should have been open circuit. It may be that this resistance is higher than the maximum that your meter can handle, though. But even if the meter just flinches momentarily away from open circuit when you touch these two points, it shows the leak.
Now: To measure it if it's out of range... You could measure the current from red to black wire with the batteries in and the switch off. Use that value and the voltage between red and black with the switch on with Ohm's law. Bob's your uncle!
Now: Return it and get a new one. This one has a leaky switch.
Ririon
Just to make 100 % sure: What you want to measure is the resistance of the switch with the switch in the off position. When you tried to do that, you took the batteries out (right). You didn't measure the resistance between the two wires (wrong), did you?. This obviously shows open circuit, since the circuit is supposed to go through the batteries...
I think you will find that the switch connects the red wire to the + pole of the last battery (the one on the right in the picture.) Take out the batteries and measure the resistivity between the red wire and this contact point. What you will find is that with the switch on, it will be very low. Your meter will probably show zero. No problem. With the switch off, it will show a value of about the size of the meter's internal resistance in voltage mode. Very high, but it should have been open circuit. It may be that this resistance is higher than the maximum that your meter can handle, though. But even if the meter just flinches momentarily away from open circuit when you touch these two points, it shows the leak.
Now: To measure it if it's out of range... You could measure the current from red to black wire with the batteries in and the switch off. Use that value and the voltage between red and black with the switch on with Ohm's law. Bob's your uncle!
Now: Return it and get a new one. This one has a leaky switch.
Ririon
Elind
Philosopher
Surely you mean V=IR ? Ohm's Law? But I'll let you off since the text of that message is still good
I appologize to Mr. Ohm.
"A" is more phonetic than "I", and a good memnonic for this law is to visualize the V over the A. Makes an X=R.
Silly maybe, but it's something from my otherwise fuzzy past.
ktesibios
Worthless Aging Hippie
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The equivalent leakage resistance across the switch which would be required to produce 1.49V across a DVM input with an input resistance of 10 Mohms is approximately 30 Mohms (assuming that the source has an open-circuit voltage of 6V).
That doesn't seem unreasonable- contamination from a variety of sources could do that. Here's a simple way to test the hypothesis that what you're detecting is switch leakage:
Get a 100 kohm resistor (any kind will do) and a couple of clip leads.
Connect the DVM to the battery holder terminals with the switch open, as you've already done. If you still obtain a similar reading, then:
Connect one clip lead to each lead of the resistor, then connect the free end of each clip lead to one of the DVM probes, so that the resistor is connected in parallel with the DVM input.
If your mysterious voltmeter reading is due to leakage currents, the DVM reading will drop dramatically (for a 30 Mohm leakage resistance and a 100 kohm load I estimate approximately 19.9 mV).
If you want to measure such a high leakage resistance, you need either a megger (an ohmmeter which uses a voltage source of a few hundred volts to measure very high resistances; commonly used in testing insulation) or a DVM which measures conductance (the reciprocal of resistance).
That doesn't seem unreasonable- contamination from a variety of sources could do that. Here's a simple way to test the hypothesis that what you're detecting is switch leakage:
Get a 100 kohm resistor (any kind will do) and a couple of clip leads.
Connect the DVM to the battery holder terminals with the switch open, as you've already done. If you still obtain a similar reading, then:
Connect one clip lead to each lead of the resistor, then connect the free end of each clip lead to one of the DVM probes, so that the resistor is connected in parallel with the DVM input.
If your mysterious voltmeter reading is due to leakage currents, the DVM reading will drop dramatically (for a 30 Mohm leakage resistance and a 100 kohm load I estimate approximately 19.9 mV).
If you want to measure such a high leakage resistance, you need either a megger (an ohmmeter which uses a voltage source of a few hundred volts to measure very high resistances; commonly used in testing insulation) or a DVM which measures conductance (the reciprocal of resistance).
Simon Bridge
Critical Thinker
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No problem. I know others have responded, but I figured I'd better respond personally. This is not indended to reflect critically on the other posts.
You can use anything you like - I've used a pencil lead, a bit of charcoal, a stick of chalk - you should be able to get a good idea of it's resistance from the meter.
If you use a commercial resistor - check the resistance with the meter first. The value painted on it is just a guide.
[aside]You realise that the 5% tolerance resistors are garanteed to be between 1% and 5% of the nominal value? (to the accuracy of the factories ohmmerters of course)[/aside]
it's been known. As stated - the fuse should blow first. However, the sensitive circuitry may get too much of a pulse before the fuse blows, or repeated abuse could cause the components to wear out. In general it is bad practise and discouraged.
It's also bad science - the circuit isn't designed to take account of the emf in the circuit. So measure resistance indirectly is always best.
Everything is real for a given value of "real"
OK - that's not helpful.
While the potential difference is really there - you don't know where it comes from. Such small values could easily be background readings.
Judging from the effect of shaking the multimeter - I'd suggest that option (1) is a strong contender there. But it could just be the difference in potential for different points in the air ... or differences between your hands.
Presumably the little (+/-) sign is flashing on the display - did you try the AC setting?
The switch will quickly charge up from the air. Most likely this effect is an artifact of your multimeter - is it perhaps a cheap-hobby multimeter?
Mine is designed for feild engineers and it still showed the kind of artifact yours does. The mistake here is to attribute the PD to the switch.
Simon Bridge
Critical Thinker
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On backgrounds and simple comparisons:
You're multimeter can measure things very accurately. Actually too accuratly for most things you want to know.
Try this: hold the multimeter probes up in the air, separated by about the width of the switch. Take several reading like that so you can get an average and a standard deviation.
The average is the background and the standard deviation is the uncertainty in the background.
Any PD within 2 uncertainties of the background is zero.
When you analyse your measurements, subtract the background from them to get the "real" PD.
Now your main readings have an uncertainty too - you could get this the same way, but you will probably need to estimate this as about half the smallest possible reading on the scale. So if the smallest possible reading is 0.1mV, the uncertainty is +/- 0.05mV.
When you subtract the background, you have to add the uncertainties. The result is the uncertainty of your corrected reading.
So if your background is 1.4+/-0.07mV and your reading is 4.5+/-0.05mV then the corrected reading is 3.1+/-0.12mV.
You're multimeter can measure things very accurately. Actually too accuratly for most things you want to know.
Try this: hold the multimeter probes up in the air, separated by about the width of the switch. Take several reading like that so you can get an average and a standard deviation.
The average is the background and the standard deviation is the uncertainty in the background.
Any PD within 2 uncertainties of the background is zero.
When you analyse your measurements, subtract the background from them to get the "real" PD.
Now your main readings have an uncertainty too - you could get this the same way, but you will probably need to estimate this as about half the smallest possible reading on the scale. So if the smallest possible reading is 0.1mV, the uncertainty is +/- 0.05mV.
When you subtract the background, you have to add the uncertainties. The result is the uncertainty of your corrected reading.
So if your background is 1.4+/-0.07mV and your reading is 4.5+/-0.05mV then the corrected reading is 3.1+/-0.12mV.
bruto
Penultimate Amazing
OK, Jyera, I've been thinking about this, and I think I have a definitive way to determine what is really going on with this box. It involves no batteries in the box, and only the ohmmeter. Here is what you do.
1. Remove the batteries from the box, and turn the switch ON.
2. Test the ohmmeter to be sure it's really working right. Use a fairly high resistance scale. You should get ~0 ohms when shorted, "1" or out of range when open. You will usually get a slight resistance reading even when you short the leads: this is the resistance of the internal fuse.
3. Now attach one ohmmeter lead to either of the output leads of the box. With the other lead, touch each of the battery contacts inside the box. One and only one contact should show a completed circuit. Make a note of which one this is, but keep testing until you have touched them all. If more than one shows anything but "open" stop right there. The box is miswired or shorted internally. If only one terminal registers, continue.
4. Now attach one ohmmeter lead to the other output lead of the box, and repeat the process above. Make sure to touch all contacts in the box again. One and only one should show a completed circuit, and again, if more than one do, the box is miswired or shorted internally. If only one lead registers, continue.
5. You now have identified the two end terminals of the battery series inside the box. Now, short together the two output leads of the box, and place the ohmmeter leads on the two identified battery terminals. When the switch is on, this should read a dead short (0 ohms). When the switch is off, this should read an open circuit. If it reads any other way, the switch is "stuffed."
This should be a pretty definitive test of the box and the switch.
1. Remove the batteries from the box, and turn the switch ON.
2. Test the ohmmeter to be sure it's really working right. Use a fairly high resistance scale. You should get ~0 ohms when shorted, "1" or out of range when open. You will usually get a slight resistance reading even when you short the leads: this is the resistance of the internal fuse.
3. Now attach one ohmmeter lead to either of the output leads of the box. With the other lead, touch each of the battery contacts inside the box. One and only one contact should show a completed circuit. Make a note of which one this is, but keep testing until you have touched them all. If more than one shows anything but "open" stop right there. The box is miswired or shorted internally. If only one terminal registers, continue.
4. Now attach one ohmmeter lead to the other output lead of the box, and repeat the process above. Make sure to touch all contacts in the box again. One and only one should show a completed circuit, and again, if more than one do, the box is miswired or shorted internally. If only one lead registers, continue.
5. You now have identified the two end terminals of the battery series inside the box. Now, short together the two output leads of the box, and place the ohmmeter leads on the two identified battery terminals. When the switch is on, this should read a dead short (0 ohms). When the switch is off, this should read an open circuit. If it reads any other way, the switch is "stuffed."
This should be a pretty definitive test of the box and the switch.