Using a multimeter as an ammeter

[Taniaz]

Homework Statement

When I connect just the multimeter (as shown in the diagram) to the power source by inserting the lead into the mA connection and change the setting to read the current on the multimeter, it reads 0.

When I change the lead connection from mA to A, the power source circuit breaker breaks.

Homework Equations

V=IR

The Attempt at a Solution

So then I tried connecting the ammeter in series with a resistor it started showing some current.

I later set up the new circuit with a rheostat, as shown in the second picture (with a diode and resistor in series with each other and another resistor parallel to it), and I tried changing the resistance using the rheostat, the current wouldn't change. Whether the diode was forward or reverse biased, it didn't seem to make a difference.

(I really hope that makes sense)

Thank you

 

[gneill]

You've described what you've done and observed, but I don't see a question.

Presumably you worked from lab instructions that provided circuit diagrams. Can you show us the schematic diagrams for your setups? It might be useful for helpers to compare the schematics with the pictures of your implementations.

 

[Merlin3189]

First: Never connect a current meter as you have shown in your first picture.
An ideal current meter is a short circuit (zero resistance) and will allow the maximum current to flow from the source. This may damage both the meter and the source.
If that were my meter, it would have blown the fuse for the mA range and that would explain why no current flows.

When you change to the A range, there is a larger fuse in the meter, so that did not blow and the supply was allowed to provide it's maximum current. Fortunately your supply seems to have overload protection and it's circuit breaker tripped.

When you use a current meter (or multimeter on current range) try to think, what is the maximum current that could possibly flow? Then start on a range greater than that. If the current shown is low enough, then you can switch to a lower range, but still greater than the maximum indicated value. So for example, if your milliamp range has a maximum value of 1999mA, and the Amp range shows 1.567A, then you can switch to the more sensitive mA range. But if the Amp range shows 2.123 A, then you cannot use the mA range. (In some cases you may find that the max current is more than your meter can handle - so don't use your meter, find one that can handle the max current, or don't do the experiment.)

Your first experiment seems to be intended to measure the short circuit current provided by the supply. It is labelled "max current # Amp", which I can't read but may be 2A, 3A or 5A? So you should have started with a meter (set on a range) that could handle that current, probably plus a decent margin (25%) for tolerances. This is not really an experiment for inexperienced people to do! Applying a short circuit (ie. amp meter) across a power supply is risky.

Your second experiment is hard to see through the tangle of wires. Perhaps. as I see gneill has now suggested, you could show the diagram of the intended circuit?
It looks to me as if the rheostat is not controlling the current through your diode. The rheostat seems to be in parallel with it, so the PSU is applied directly across the diode circuit. Also, it would help to know the values of your resistors.

 

[Taniaz]

Thank you for your replies. It makes a lot of sense now. I've attached the schematic for you to have a look at.

The experiment asks us to use resistors of 100 and 220 ohms but we only have 4.7 kilo-ohms and 5, 10, 20 and 50 ohms.

So we used the 25 ohms resistor in series with the diode and the 50 ohms resistor in parallel to the two. We could attach these in series to give the 100 and 220 ohms they want but the connection was getting much crazier that way.

My question basically was what am I doing wrong with the rheostat and the multimeter which is acting as an ammeter?

 

[gneill]

My question basically was what am I doing wrong with the rheostat and the multimeter which is acting as an ammeter?

@Merlin3189 covered the use of ammeters and why it's not a good idea to place one directly across a voltage source.

For the rheostat, you need to understand the connections on the device and match them to your circuit diagram. You've only used two of the three available connections, whereas the circuit requires the use of all three.

The rheostat consists of two main components, the cylindrical coil of resistive wire and the sliding contact. The connectors I've labelled A and B connect to either end of the coil (resistor) while connector C connects to the slider. I've placed a schematic representation is below the image.

The circuit in your lab drawing shows both ends of the "resistor" part of the rheostat connected to the power supply, while the image of your setup shows the power supply connected to one end of the coil and the slider. Effectively you've built this:

As you can see, the rheostat is not controlling the voltage delivered to the load.

 

[Taniaz]

I thought connect all three connections of the rheostat then it behaves as a potential divider. So it this particular experiment, it has to behave as a potential divider?

Is this my only mistake?

 

[gneill]

I thought connect all three connections of the rheostat then it behaves as a potential divider. So it this particular experiment, it has to behave as a potential divider?

From what I can see from the provided circuit diagram it's the only connection error.

Is this my only mistake?

I'd be concerned about using resistance values that are much smaller than those recommended, particularly the one in series with the diode: It's meant to protect the diode from excessive current. Without seeing the rest of the lab instructions it's hard to say if you'll run into any other issues due to the change in resistor values.

 

[Taniaz]

We're trying to get the resistors they recommended. When we used the 4.7 kilo ohms resistance, the multimeter showed us a current of 0 (it was the only other option we had).

 

[gneill]

We're trying to get the resistors they recommended. When we used the 4.7 kilo ohms resistance, the multimeter showed us a current of 0 (it was the only other option we had).

If you have multiples of the smaller resistor values, use series connections to create the desired values. Yes there will be a lot of connections, but you can build these separately first and check their values with the multimeter.

 

[Taniaz]

Yes we did that too as mentioned but the circuit just looked even crazier and harder to manage since we only had one 50 ohm resistor and 20 ohms and quite a few smaller ones such as 5 and 10. Thank you, we'll try it out tomorrow and give you the update regarding what happens.

So we have to connect A and B to the source and C to the multimeter

 

[gneill]

So we have to connect A and B to the source and C to the multimeter

Correct.

 

[Taniaz]

Can you just verify this for us, when we want to use the rheostat as a variable resistor we only make the two connections but when we want to use it as a potential divider we have to use all three connections?

Thank you for your help

 

[gneill]

Can you just verify this for us, when we want to use the rheostat as a variable resistor we only make the two connections but when we want to use it as a potential divider we have to use all three connections?

That's true.

 

[Taniaz]

Thank you, do you mind if I ask you another lab question regarding capacitors? It's a really small question regarding the time versus resistance graph when a capacitor discharges so it doesn't make sense making a separate post for it.

 

[gneill]

Thank you, do you mind if I ask you another lab question regarding capacitors? It's a really small question regarding the time versus resistance graph when a capacitor discharges so it doesn't make sense making a separate post for it.

Actually, it would be preferable to make a separate post. If the question is not described in the initial post of the thread, a new post is required.

 

[Taniaz]

Alright great, thank you!

 

[Taniaz]

Hi, so we tried it today and it works! But there is one slight issue, when we connect the diode as reverse biased we shouldn't pick up changing current. When we slide the rheostat, the current changes from 0-2.4 mA for reverse biased! And 0-4.2 mA for forward biased.

And the even bigger confusion is when we connect our diode as forward biased it shows the one with the smaller range.

 

[TomHart]

It wasn't clear to me what the values of the resistors are within "Component M". But you have two parallel paths within Component M. One contains a diode in series with a resistor, and the other path contains a resistor. Isn't it true that, even if the diode is reverse-biased, the current through the single-resistor path will vary as you vary the rheostat? But like I said, I don't know what the resistor values are in Component M.

 

[Taniaz]

Ah yes that makes sense, the resistors they recommended are 100 and 220 ohms but we don't have those so we are just testing with a 10ohms and 20 ohms, we don't even have enough of those otherwise we could have attached them in series to get the recommended resistance. We've tried to get them to order some so we're just waiting now.

So when the diode was actually forward biased, it showed the smaller range values and when it was reverse biased, it showed the larger range values.

 

[TomHart]

What is your DC voltage set to?

Edit: And as @gneill pointed out, you should be careful of using small resistor values - such as what you are using. It sounds like there is a problem if the maximum current is higher for the reverse-biased case.

 

[Taniaz]

The DC voltage used is 6V. So what do you suggest we use if we don't have 100 and 220 for the time being?

Secondly, they asked us to record the voltage across the component M when the diode is forward biased by setting the current to 16 mA.

When we set our dial to 200 mA range we get 1.6 and when we change the range to 10, it shows 0.16? We can't seem to get a voltage for 16 mA

 

[Taniaz]

Ok so we have all the values they required with our readings and the maths works out.

Could you just re-explain your point regarding why when the diode is forward biased the current doesn't change much but when it is reverse biased the current changes much more?

Thank you

 

[TomHart]

If you measure 0.16 A when the scale is set to 10 A, then you should be able to vary the rheostat to drop it down to 16 mA.
But, if you are measuring maximum currents of 2.4 mA and 4.2 mA (forward- and reverse-biased) - that is, with the rheostat wiper set to maximize the voltage to Component M - then you definitely have a problem. When you switch current scales on the multimeter, are you required to change the plug-in location on the meter? Is there possibly a fuse blown on the multimeter input?

What is the full-scale resistance of the rheostat? Can you measure it from one end to the other (not the wiper), or is it specified on a label?
If you set the wiper of the rheostat to maximize the voltage to Component M, and the DC voltage source is set to 6 V, you should be able to measure 6 V (or close to it) across Component M. And with 6 V across Component M in the forward-biased direction, with a 10 ohm and 20 ohm in parallel, the current should be close to 1 Amp through your ammeter. So if you are only measuring a few milliamps, something is wrong.

Maybe your DC supply is current limiting. Is there a current limit adjustment on the DC voltage source?
But I'm still concerned that the 10 and 20 ohm resistors are just too low of a resistance for this case.

 

[TomHart]

Could you just re-explain your point regarding why when the diode is forward biased the current doesn't change much but when it is reverse biased the current changes much more?

Forward-biased or reverse-biased, the current should change significantly. But forward biased, the maximum current should be higher than in the reverse-biased case.

The diode just basically acts like a switch. When it is forward biased, the switch is closed, allowing current to flow through the resistor that is in series with the diode. When the diode is reverse-biased, the switch is open, preventing any current from flowing through the resistor that is in series with the diode. But in both cases, current is still free to flow through the other path that only has a single resistor (no diode). So when forward-biased, the maximum current should be higher because you basically have 2 resistors in parallel.

 

[Taniaz]

If you measure 0.16 A when the scale is set to 10 A, then you should be able to vary the rheostat to drop it down to 16 mA.
But, if you are measuring maximum currents of 2.4 mA and 4.2 mA (forward- and reverse-biased) - that is, with the rheostat wiper set to maximize the voltage to Component M - then you definitely have a problem. When you switch current scales on the multimeter, are you required to change the plug-in location on the meter? Is there possibly a fuse blown on the multimeter input?

What is the full-scale resistance of the rheostat? Can you measure it from one end to the other (not the wiper), or is it specified on a label?
If you set the wiper of the rheostat to maximize the voltage to Component M, and the DC voltage source is set to 6 V, you should be able to measure 6 V (or close to it) across Component M. And with 6 V across Component M in the forward-biased direction, with a 10 ohm and 20 ohm in parallel, the current should be close to 1 Amp through your ammeter. So if you are only measuring a few milliamps, something is wrong.

Maybe your DC supply is current limiting. Is there a current limit adjustment on the DC voltage source?
But I'm still concerned that the 10 and 20 ohm resistors are just too low of a resistance for this case.

Sorry the range was from 0- 0.24 A and 0-0.42 A not 2.4 mA and 4.2 mA.
The multimeter only gives us readings when it is plugged into the 10A max location not when we plug it into VmAΩ location.
I don't think there's a fuse blown because it is giving us a variation in current and when we use it as a voltmeter it does give us the voltage as expected from 0-6 V.

 

[Taniaz]

Forward-biased or reverse-biased, the current should change significantly. But forward biased, the maximum current should be higher than in the reverse-biased case.

The diode just basically acts like a switch. When it is forward biased, the switch is closed, allowing current to flow through the resistor that is in series with the diode. When the diode is reverse-biased, the switch is open, preventing any current from flowing through the resistor that is in series with the diode. But in both cases, current is still free to flow through the other path that only has a single resistor (no diode). So when forward-biased, the maximum current should be higher because you basically have 2 resistors in parallel.

This is what we expected too. That's the problem, we aren't observing this.

 

[Taniaz]

Actually no, sorry, I was looking at it the other way around! My bad, works perfectly and the maths works out too! Thank you once again!

The anode should be connected to the positive terminal of the power supply yes? (facepalm moments so sorry!)

 

[TomHart]

I was looking at the current variations and was wondering if you might have just been confused with the polarity (forward vs. reverse) - because it is greater in one case than the other. However, the difference between 0.24 and 0.42 is not as much as I would expect. <-- Edit: (Actually, the difference depends on which of the 2 resistors is in series with the diode.) But all is well if you are confident with the outcome.

 

[Taniaz]

I wasn't too confident about it at first but then we started answering their questions. I'll just lay them out for you.

i) Connect the positive terminal of the battery and set the current to 16 mA. What is the voltage V1 across the component M?

We went along with 0.16 A instead of 16 mA and we got a voltage of 3.16 V.

ii) Reverse the component M and set the current to 32 mA. What is the voltage V2 across the component M?

We went along with 0.32 A and we got a voltage of 2.52 V.

iii) Component M contains the items shown in the diagram shown below (the schematic attached previously with the diode in series with one resistor and in parallel to another resistor). Use part (i) to find the resistance of the resistor R'.

This was the reverse biased connection so by theory if we assume that the diode did not allow any current to pass through then all the current went to the resistor in parallel and so R' = V/I = 3.16 V / 0.16 A = 19.75 ohms which is close since that resistor was 20 ohms.

iv) Assume the voltage drop across the diode is 0.60 V when forward biased. For the arrangement in part (ii) when the upper and lower part of component are conducting, find the following:

-current flowing through resistor R'

I = V/R = 2.52 / 19.75 = 0.127 A

- current flowing through resistor R

I = 0.32 A - 0.127 A = 0.193 A

-the resistance of the resistor R.

The voltage across R is 2.52 V -0.60 V=1.92 V and therefore R = 1.92 V / 0.193 A = 9.9 ohms which is very close to 10 ohms (the original value we were using for R)

 

[gneill]

Sorry the range was from 0- 0.24 A and 0-0.42 A not 2.4 mA and 4.2 mA.
The multimeter only gives us readings when it is plugged into the 10A max location not when we plug it into VmAΩ location.
I don't think there's a fuse blown because it is giving us a variation in current and when we use it as a voltmeter it does give us the voltage as expected from 0-6 V.

The fuse may only affect the current measuring operation depending on how the multimeter is designed. You can test the meter quite easily as follows. I believe you said that you have a 4.7 kΩ resistor, correct? Place that in series with one of the meter leads and connect to the 6 V power source. The current reading should be about 1.3 mA: $I = V/R = 6/4700 = 1.277~mA$.

 

[TomHart]

I wasn't too confident about it at first but then we started answering their questions. I'll just lay them out for you.

i) Connect the positive terminal of the battery and set the current to 16 mA. What is the voltage V1 across the component M?

We went along with 0.16 A instead of 16 mA and we got a voltage of 3.16 V.

ii) Reverse the component M and set the current to 32 mA. What is the voltage V2 across the component M?

We went along with 0.32 A and we got a voltage of 2.52 V.

iii) Component M contains the items shown in the diagram shown below (the schematic attached previously with the diode in series with one resistor and in parallel to another resistor). Use part (i) to find the resistance of the resistor R'.

This was the reverse biased connection so by theory if we assume that the diode did not allow any current to pass through then all the current went to the resistor in parallel and so R' = V/I = 3.16 V / 0.16 A = 19.75 ohms which is close since that resistor was 20 ohms.

iv) Assume the voltage drop across the diode is 0.60 V when forward biased. For the arrangement in part (ii) when the upper and lower part of component are conducting, find the following:

-current flowing through resistor R'

I = V/R = 2.52 / 19.75 = 0.127 A

- current flowing through resistor R

I = 0.32 A - 0.127 A = 0.193 A

-the resistance of the resistor R.

The voltage across R is 2.52 V -0.60 V=1.92 V and therefore R = 1.92 V / 0.193 A = 9.9 ohms which is very close to 10 ohms (the original value we were using for R)

Looks great - exactly what you would expect!

 

[Taniaz]

Looks great - exactly what you would expect!

Thank you for your help!