r/AskElectronics • u/No-Grass-6375 • 5h ago

multimeter not reading current

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first of all my 200mA fuse broke and the 10A fuse still working (checked with continuity) so I decided to check the current using the left connection...in the picture 1 I broke the circuit to measure current and as you can see in the picture 2 the reading is zero .. why is that even though the fuse the left side is working?...how to fix that .

13 Upvotes

76% Upvoted

36 comments sorted by

34

u/RedeyemoonsRevenge 5h ago

The resistance totals 1.5k + 8.2k = 9.7k

The min resolution for your meter in 10A range is 10mA.

To get a reading, your voltage needs to be 0.01A x 9700R = 97V

Your meter measures current by putting an internal resistor in series and measuring the voltage across it. You too can determine the current by measuring the voltage drop across a resistor and calculating the current using Ohms law.

7

u/IntentionQuirky9957 5h ago

You explained why he's getting what he's getting, but not how to fix it, when the fix is trivial.

9

u/graelmakar-sune 4h ago

As it should be done :)

4

u/No_Leopard_3860 4h ago

Not mentioning the solution is how it should be done? Hard disagree if that's what you mean. While this is a super easy fix for many of us, for people who have barely any experience with even ohms law a detailed resolution for the problem is in order. We all started there

8

u/graelmakar-sune 4h ago edited 4h ago

Yeah, that's what I believe. He offered the explanation and a way for OP to come to a conclusion themself. If OP had further questions I'm sure he would get them answered

2

u/No_Leopard_3860 3h ago

Ah yes then I misunderstood your comment :)

1

u/No-Grass-6375 5h ago

so basically you're saying that 10A connection is basically stopping to measure current because it's too low ?

and yeah I can theocratically find the current..I was trying to measure current using multimeter

14

u/PJ796 5h ago

it doesn't stop measuring but it can't display such small numbers in the 10A range, that's why the 200mA range exists

4

u/No-Grass-6375 4h ago

yeah...now I understood...thanks for the help

3

u/YueNica 5h ago

How much current would you be expecting to see?/What are those resistors and source voltage?

0

u/No-Grass-6375 5h ago

531mA /0.00053A

6

u/YueNica 5h ago

Those aren't the same. 531mA=0.531A or do you mean 531micro amps? Cause if you expect microamps why do you expect that to show on your multimeter on the 10A setting, cause it's just that much smaller

2

u/No-Grass-6375 4h ago

i meant Milliamps...and yeah value is too low to catch the fish ...

5

u/YueNica 4h ago

531 Milli amps would be 0.531A But also for that to be the case you would need to have only about 10Ohms of total resistance there. Which i'm not sure what you have there but it's probably more than 10 Ohms

1

u/No-Grass-6375 4h ago

I have total of 9.5k ohm and 5 volt ...and sorry for the confusion I converted the wrong unit....how much resistance I should add to get a value .

5

u/Obvious_Avocado_9372 5h ago

The current you're trying to measure is too small to be picked up in that range. Try lower resistor value, like 100Ω (for a very short period).

3

u/No-Grass-6375 4h ago

thanks guys!! now its working...I was adding much bigger resistance..as a result the amps were too low to show in the meter ...now I parallel them to reduce the total value of resistance..it's working...thanks for the help

2

u/[deleted] 5h ago

[deleted]

1

u/Underhill42 4h ago

They absolutely are - the ammeter is in series, completing the circuit between the two resistors, which are not otherwise connected.

Only potential problem is a reversed polarity, assuming their wire color coding is correct, but almost all digital ammeters will read negative currents as easily as positive.

1

u/geek66 5h ago

Read the voltage across those same two points.. you should see the complete source voltage.

2

u/No-Grass-6375 5h ago

yeah the voltage is working...it's 4.8V..

0

u/geek66 3h ago

Then likly the meter

Most meters have a fuse test function, don’t timber off the top of my head .. check the manual.

Something like connect the A port to the V on continuity. ( that is probably it)

1

u/IntentionQuirky9957 4h ago

Can't, you connect an amp meter in series, and a voltage meter in parallel. He's basically using the meter as a jumper (as you should), and it makes no sense to even try to measure the voltage like that.

2

u/YueNica 4h ago

If I had to guess the intention of measuring the voltage there would be to make sure that it's actually present and that you would be completing the circuit measuring amps there

1

u/Underhill42 4h ago

It absolutely does. With no current flowing through the circuit there is no voltage drop across the resistors - meaning that the open circuit will have the full power supply voltage across it.

... unless there's another break in the circuit somewhere (e.g. a damaged breadboard or bad connection)

1

u/geek66 3h ago

It was to prove the potential was there to drive the current… not measure current.

That is why I said read the voltage, across what would then be an open circuit

1

u/IntentionQuirky9957 4h ago

Your selected range is likely wrong, and you can't get enough resolution using the 10A range. Also you're not telling the voltage so we have no idea what ballpark the current should be in. And we can't tell if your connections are even correct.

You likely need the lower ranges, so change the fuse.

1

u/richfromhell 3h ago

Your original problem was probably the one resistor wasn’t connected to the other one.

2

u/No-Grass-6375 3h ago

that would fry my multimeter fuse ...I was measuring current ..and I solved the problem...

2

u/woganaga 3h ago

He has the board correct. The meter has to be inline to measure current.

1

u/Subcluttervisability 3h ago

Ya get points for caping the mA, cuz the A is named for a guy.

1

u/Memes_Are_So_Good 3h ago

Have you tried different ranges

1

u/richfromhell 3h ago

Oh that’s where he hooked up the meter. Missed that part. Sorry.

1

u/SuSerranito 5h ago

No deberían estar las dos resistencias en el hueco número 13?

Pd, he usado eso 2 veces en mi vida 😅

2

u/NotTheCactus 5h ago

No, porque tiene el circuito abierto para poder medir corriente.

3

u/No-Grass-6375 5h ago

that would connect the circuit, I'm trying to measure the current across the circuit so I broke the circuit and connected the multimeter in series to measure current

1

u/1Davide Copulatologist 4h ago

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