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I have a \$1.5\Omega\$ resistor and \$8.84V\$ battery. My multimeter tip's resistance is \$0.3\Omega\$.

I've tried to compare the calculated voltage drop (which is \$8.84V\$) with the measured value.

What I've tried

I've connected the resistor to the battery and then connected the tips to the battery terminal to measure the voltage between terminal. (The time was short. About 3 seconds)

The value should be \$0V\$ but it was \$0.34V\$ (means the voltage drop is \$8.5\$).

Why is there a difference?

enter image description here enter image description here

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  • \$\begingroup\$ Why do you think the voltage between battery terminals should be $0$?? \$\endgroup\$ – Ali Aug 31 '13 at 12:31
  • \$\begingroup\$ There is a single factor which is resistor in the circuit. The input voltage on the resistor is 9[V]. From the law of Ohm, V=IR, the V is 9[V] which leads the voltage drop. \$\endgroup\$ – user28936 Aug 31 '13 at 12:41
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    \$\begingroup\$ The battery has its own internal resistance. \$\endgroup\$ – Mike Dunlavey Aug 31 '13 at 15:26
  • \$\begingroup\$ @Qmechanic Confirmed. \$\endgroup\$ – user28936 Sep 1 '13 at 9:22
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I don't quite understand why the calculated voltage drop is 8.84V - every battery has nonzero internal resistance, so it looks like what you measured was the voltage drop on this internal resistance.

EDIT (09/01/2013): Your schematics seems to confirm that you measured the voltage drop on the internal resistance of your battery. I also suspect that your "multimeter tip's resistance" of 0.3Ω is rather irrelevant, as a decent voltmeter itself should have a much greater resistance. However, your voltage drop of 0.34V seems very low, as the internal resistance of the battery should be about 1.7Ω (http://www.alliedelec.com/images/products/datasheets/bm/DURACELL/70149225.pdf ), so the current in the circuit should be greater than, say, 2.5A, and the voltage drop on the battery should be at least 4V. Maybe I am missing something?

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  • \$\begingroup\$ -1: This answer is incorrect; the voltmeter does not read the voltage across the internal resistance of the battery. \$\endgroup\$ – Alfred Centauri Sep 1 '13 at 10:03
  • \$\begingroup\$ I stand corrected - forgot about battery voltage:-) \$\endgroup\$ – akhmeteli Sep 1 '13 at 10:08
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you are measuring the EMF of the battery less the voltage drop across the battery's internal resistance, OR the voltage drop across the 1.5ohm, which will not be equal to battery EMF because of said internal resistance. i.e. +0.34= +8.84 - (+Vir), as diagram below.enter image description here

your result shows that the internal resistance of the battery is 8.5/0.34*1.5=37.5ohm. which is much higher than one would expect. it is usually 1-2ohm. either the battery is much less than +8.84V or the load resistor is more than 1.5ohm

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  • \$\begingroup\$ EMF = Electromotive Force = Voltage \$\endgroup\$ – JYelton Sep 1 '13 at 11:27
  • \$\begingroup\$ I'm not sure about this, but the internal resistance may increase as the battery heats up. Which a 9 V will do if you try to pull a couple of amps out of it. \$\endgroup\$ – The Photon Sep 1 '13 at 16:12
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If the battery, resistor, and voltmeter are connected in parallel, you should read the only voltage present in the circuit (remember, parallel connected circuit elements share the same voltage.)

If you are measuring just 0.34V, the battery is probably exhausted and can only supply a relatively small current:

\$I_R = \dfrac{0.34V}{1.5\Omega} = 227mA\$

Now, if the battery, resistor, and voltmeter are connected in series, you should read the open circuit battery voltage of 8.84V since there should be essentially zero current (remember, an ideal voltmeter is an open circuit).

It would be helpful if you added a circuit diagram so that we can see exactly what you are trying to measure.


From the schematic, you should read the battery open circuit voltage when the switch is open. A fresh battery should read about 9.35V. Since you're reading 8.84V, the battery is at or near discharge.

With the switch closed, a fresh battery will drive a significant current, several amps, through the external resistance as it rather rapidly discharges. Since your battery is supply much less than that, it is already discharged.

Here is a link to an experiment to determine the short circuit current of a 9V battery that I think you may find informative:

http://friedrichengineering.com/web_documents/9volt%20Battery.pdf

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  • \$\begingroup\$ I've added the images. \$\endgroup\$ – user28936 Sep 1 '13 at 4:42

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