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I have a very simple circuit: a 9V battery is connected to a 1500ohm resistor, which is connected in series to the multimeter and back to the battery.

schematic

simulate this circuit – Schematic created using CircuitLab

According to Ohm's Law, I should get:

I = V/R = 9.15V (measured) / 1496ohm (measured) = ~0.00612A = 6,12mA

I set my multimeter to 200mA reading and I get 30.2mA. Then I set my multimeter to 20mA reading and I get 7.89mA.

Why do I get two different amperages when I set a different limit? Why are the readings so off compared to the theoretical amperage? I just don't get it.

I don't have another multimeter right now so I can't verify whether my unit is faulty or I'm doing something wrong.

The same issue occurs when I add a led diode to the circuit (which draws approximately 2V, so I consider a difference of voltage of 7V instead of 9V): the readings are off.

I just started with electronics, so maybe this is a stupid question, but I can't wrap my head around it.

Thanks.

Edit: I attach two images showing the readings. I switched batteries several times, both the one powering the multimeter and the one used for the circuit. 200mA limit

20mA limit

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  • \$\begingroup\$ Change your DVM battery and make sure you are measuring DC and not AC. \$\endgroup\$
    – Andy aka
    Nov 22 '20 at 13:13
  • \$\begingroup\$ Your test does indicate something is wrong. Another test to try...reverse multimeter leads so that you get a negative reading: its magnitude should be the same as a positive reading. Be aware that current shunt resistors inside your multimeter are vulnerable to burnout. An internal fuse should provide some protection. \$\endgroup\$
    – glen_geek
    Nov 22 '20 at 14:20
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Most likely your multimeter is faulty, inaccurate, or simply low on batteries.

There's no way to get more than about 6mA flowing via the resistor from the 9V battery, unless you have shorted the battery, but then it would be more than 30mA.

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enter image description here

Figure 1. A 200 mV meter module. Image source: Farnell.

Most digital multimeters use a voltmeter module for the display. The most common are the 199.9 mV type and so the full range reading on any of the amps scales requires that 200 mV is dropped across the shunt. That means that -

  • The 200 mA range shunt will be \$ R = \frac V I = {200m}{200m} = 1 \ \Omega \$.
  • The 20 mA range shunt will be 10 Ω.

Clearly they're not going to make much of a difference when you have 1.5 kΩ in circuit already.

Please post a photo of your setup showing the meter and lead connections in use. Link to the manual if possible.

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  • \$\begingroup\$ Thanks, I added the photos to the OP showing both readings. Hope that helps. I used two alligator cables but the issue is still there even without them. \$\endgroup\$
    – skaine
    Nov 22 '20 at 13:42

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