Errors in Analogue & Digital Multimeters: Why?

[Tasell]

I've a circuit with a power supply of 1.95V and a resistor 100ohms. To measure the current through the resistor, I use an analogue multimeter. This turns out to be 1.9x10^-4 A. But, from the calculation using the formula I=V/R, I should be 1.95x10^-2. Shouldn't an analogue multimeter be accurate to measure current due to its low resistance? Or, is 1.95V too high for the multimeter?

And, to calculate the internal resistance of the AMM, I use a digital multimeter to measure the voltage dropped, which turns out to be 0.08V. Then, plugging it into the equation r=V/I=0.08V/1.9x10^-4 = 4.21ohm. Again, from the theoretical calculation, r=(V/I)-R=(1.95/1.9x10^-4)-100 = 2.63ohm. This is very different from the measured value. Can anyone explain to me why this is so?

 

[Georgepowell]

I've a circuit with a power supply of 1.95V and a resistor 100ohms. To measure the current through the resistor, I use an analogue multimeter. This turns out to be 1.9x10^-4 A.

Well something must be wrong... It may be that you have a faulty wire, a faulty connection somewhere, or a faulty multimeter. You may have also set up the circuit wrong? did you put the multimeter in series or in parallel?

 

[Gnosis]

I've a circuit with a power supply of 1.95V and a resistor 100ohms. To measure the current through the resistor, I use an analogue multimeter. This turns out to be 1.9x10^-4 A. But, from the calculation using the formula I=V/R, I should be 1.95x10^-2. Shouldn't an analogue multimeter be accurate to measure current due to its low resistance? Or, is 1.95V too high for the multimeter?

And, to calculate the internal resistance of the AMM, I use a digital multimeter to measure the voltage dropped, which turns out to be 0.08V. Then, plugging it into the equation r=V/I=0.08V/1.9x10^-4 = 4.21ohm. Again, from the theoretical calculation, r=(V/I)-R=(1.95/1.9x10^-4)-100 = 2.63ohm. This is very different from the measured value. Can anyone explain to me why this is so?

Right off the bat, it sounds like you're using a 10K ohm resistor rather than a 100 ohm resistor. That would move your decimal point 2 places as it did.

The color code for a 100 ohm resistor is: brown, black, brown

HOWEVER, many browns and reds of resistors are VERY close in appearance especially when the color bands are particularly narrow, so a 10K ohm resistor (brown, black, red) can fairly easily appear to be a brown, black, brown. You wouldn't be the first to make this visual error and you won't be the last.

Ohmmeter the 100 ohm resistor in question.

 

[dlgoff]

Ohmmeter the 100 ohm resistor in question.

Yes. Also note the tolerance of your resistor. If it's a 5% resistor, a 100 ohm resistor will probably have a value from 95 to 105 ohm.

 

[natey]

I've a circuit with a power supply of 1.95V and a resistor 100ohms. To measure the current through the resistor, I use an analogue multimeter. This turns out to be 1.9x10^-4 A. But, from the calculation using the formula I=V/R, I should be 1.95x10^-2. Shouldn't an analogue multimeter be accurate to measure current due to its low resistance? Or, is 1.95V too high for the multimeter?

And, to calculate the internal resistance of the AMM, I use a digital multimeter to measure the voltage dropped, which turns out to be 0.08V. Then, plugging it into the equation r=V/I=0.08V/1.9x10^-4 = 4.21ohm. Again, from the theoretical calculation, r=(V/I)-R=(1.95/1.9x10^-4)-100 = 2.63ohm. This is very different from the measured value. Can anyone explain to me why this is so?

yes sounds suspiciously like a 10k resistor to me, check its resistance with the multimeter, measuring such small voltages and currents is never going to be hugely accurate, it all depends on the accuracy of the multimeter, has it got a uA, mA or A setting, some are better than others, also has it been calibrated?