Ohm's Law doesn't work?

Thread Starter

Yaşar Arabacı

Joined Nov 11, 2014
49
Hi,

I have been reading http://www.allaboutcircuits.com/vol_1/chpt_10/3.html and wanted to try it on my own. After setting up my circuit, I tried to read the voltage drop and current trough each resistor. Interestingly, my reading doesn't conform to Ohm's Law.

See the image with my readings and schematic of the circuit: http://i.imgur.com/akIiGAc.jpg

For example, one of the 1000 ohm resistor reads voltage drop of 3.63V and current of 8.84mA which is weird because 8.84mA x 1000ohm = 8.84V which is different from 3.63V.

Which of the following is true?

a) My understanding of Ohm's Law is wrong
b) Ohm's Law is wrong
c) My resistors's resistance changes with voltage (I don't think they are supposed to do that?)
d) My multimeter is broken (I doubt it, I just bought it yesterday)
e) Go home physics, you are drunk?
 

alfacliff

Joined Dec 13, 2013
2,458
ohms law is NOT wrong.
placing the meter in series with the resistor which you wold have to do to measure current adds the resistance of the meter.
 

WBahn

Joined Mar 31, 2012
30,076
My guess is that the problem is that you aren't making your measurements properly.

How, exactly, are you making your current measurements?
 

Thread Starter

Yaşar Arabacı

Joined Nov 11, 2014
49
I am setting up my multimeter for DCA 20m mode because I don't think I will read more than 20mA's and I am touching the resistor's each metal cable with black and red cables of multimeter, and I wait a little until the numbers on my multimeter stabilizes.
 

WBahn

Joined Mar 31, 2012
30,076
That's what I thought. You are making perhaps one of the most common mistakes for newbies. In order to measure current (with this type of meter), you must place the meter in series with the current you wish to measure. So for each resistor you will need to disconnect one lead of the resistor and put one lead of the meter in it's place. Then connect the other lead of the meter to the disconnected lead of the resistor.

Also, voltages and currents are polarized quantities, so just giving a number is not sufficient. You need to also indicate which side of the voltage is positive and which direction the current is flowing. With most meters, the positive side of a positive reading voltage measurement is at the red probe and a positive current reading means that the current is traveling through the meter from the red to the black probe.
 

MaxHeadRoom

Joined Jul 18, 2013
28,698
Ohms Law.
Following on what others have noted, you could also take a current reading by reading the voltage drop across a known resistance and calculate the current using ohms law, in this case the meter is on voltage and should present as little impedance as possible to the circuit, ideally have no resistance, otherwise the meter is placing its resistance in parallel to the measured resistor.
Max.
 

wayneh

Joined Sep 9, 2010
17,498
You are making perhaps one of the most common mistakes for newbies.
It is SO common that we can almost take it for granted. The OP is lucky he didn't measure the current from the battery this way - he'd be looking for a new fuse for his meter like so many others before him. I admit it, I've done it. I like to think I won't do it again, but I have spare fuses. :rolleyes:
 

WBahn

Joined Mar 31, 2012
30,076
It is SO common that we can almost take it for granted. The OP is lucky he didn't measure the current from the battery this way - he'd be looking for a new fuse for his meter like so many others before him. I admit it, I've done it. I like to think I won't do it again, but I have spare fuses. :rolleyes:
I never made this mistake intentionally, but I've done it a couple of time inadvertently. I bought a new Micronta meter back in the 1989 time frame (the Radio Shack brand was a decent meter back then) and was measuring the current draw on a bunch of things in my room to estimate how much electricity I was using since my landlady was concerned because of all my electronics. I was using the 10A (which is unfused) range and for some reason I decided to verify the voltage and I forgot to switch from A to V. After replacing the old 15 A Edison fuse in the basement fuse box I figured I had just thoroughly toasted that meter. But it still works to this day.

But I have to admit that I very seldom ever take current measurements in part because I don't like the inaccuracy due to the series resistance and in part because I don't think I have any meter except my newest cheapie that has a good fuse in it.
 

wayneh

Joined Sep 9, 2010
17,498
...I very seldom ever take current measurements in part because I don't like the inaccuracy due to the series resistance.
Ditto. I wish they'd let you know what the approximate series resistance is. It's low enough that I don't think I can measure it with another meter. I guess I could use one meter to measure ∆V across another meter while measuring current to a headlight or such.
 

WBahn

Joined Mar 31, 2012
30,076
Ditto. I wish they'd let you know what the approximate series resistance is. It's low enough that I don't think I can measure it with another meter. I guess I could use one meter to measure ∆V across another meter while measuring current to a headlight or such.
The specs on the meter should give you that information. Even my cheapo meters give me some indication. On the lower current ranges the resistance can be pretty significant.
 

wayneh

Joined Sep 9, 2010
17,498
The specs on the meter should give you that information. Even my cheapo meters give me some indication. On the lower current ranges the resistance can be pretty significant.
I have the infamous Cen-Tech meter. Several, actually. They're free at Harbor Freight, so I grab one anytime I'm in the neighborhood. The manual does not address the shunt resistance and I haven't been able to find any reference on it.

There are quite a few references to its 1MΩ input impedance (on all scales) for voltage measurements, which I've confirmed myself for the 20V scale.
 
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