# Finding the internal resistance of battery

## Homework Statement

I measured a battery rated at 1.5V(ε) and get 0.9V out and want to calculate the internal resistance (r). So, I model the battery as an ideal voltage source and a resistor in parallel. I know the voltmeter has a resistance of 1kΩ (R). How do I calculate the internal resistance?

V=IR, ∑V = 0

## The Attempt at a Solution

Subtracting 1.5-0.9 gives me 0.6V, this is the voltage lost across the internal resistor, so ΔV = 0.6V = Ir
I also know that the sum of the voltages around the circuit is zero, but I'm not sure if this means there is a voltage drop across the voltmeter resistor of 0.9V and across the internal "resistor" of 0.6V or if the internal voltage drop is equal to the voltage drop across the voltmeter. But, what I have done algebraically is this:

$$\epsilon = \Delta V_I + \Delta V_{vm} = I_ir+I_{vm}R$$

This is where I get stuck, I know what ##\epsilon## is and I know what the ##\Delta V##'s are, and I know what R is, but I don't know the current for the voltmeter or the current across the internal resistor, so I'm stuck. I have tried to use Ohm's law to express the currents as ##\frac{\Delta V_{i/vm}}{r/R}## but it just ends up being circular logic and I get some form of ##R_i = R_i##.

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I model the battery as an ideal voltage source and a resistor in parallel.
The resistor should be in series with the ideal source.

rmiller70015
The resistor should be in series with the ideal source.
I screwed up I meant the resistor is in series with voltage source and the voltmeter resistor is in parallel with the internal resistor.

It helps to make a quick sketch of the circuit. I think the voltmeter, ideal source and Thevenin resistor are all in series.

rmiller70015
If I were to physically connect the voltmeter to the circuit wouldn't it be on either side of the resistor, putting the voltmeter resistor in parallel with the internal resistor, or would the resistors be in series as a consequence of the internal resistor being theoretical?

I hope I understand the problem set up correctly. The voltmeter would not be on either side of the resistor. The black meter lead would be on the negative battery terminal, the red meter lead would be on the free end of the resistor (the end that's not connected to the positive battery terminal).

rmiller70015
I hope I understand the problem set up correctly. The voltmeter would not be on either side of the resistor. The black meter lead would be on the negative battery terminal, the red meter lead would be on the free end of the resistor (the end that's not connected to the positive battery terminal).
Yes, that would be correct, so it would be in series then. Thank you very much.

Hi I would hope the multimeter has a much greater internal resistance in the voltage measurement mode than 1K. At a 1K internal resistance the meter would likely load the circuit down and give you a false reading. My fluke 87v has a 10 Mega ohm internal resistance. It does not load down most of my electronic projects

gneill
Mentor
Hi I would hope the multimeter has a much greater internal resistance in the voltage measurement mode than 1K. At a 1K internal resistance the meter would likely load the circuit down and give you a false reading. My fluke 87v has a 10 Mega ohm internal resistance. It does not load down most of my electronic projects
Don't confuse real-life scenarios with academic exercises

If the given problem says that the meter has a 1 KΩ resistance, you've got to take that as a given.

Thank you for pointing that out. I will look more closely at how the question is being asked in the future. It’s been a while since I have been in college