# Internal Resistances

1. Sep 11, 2008

### bengaltiger14

1. The problem statement, all variables and given/known data

Calculate internal resistance of meter. There is a 10V power supply with a variable resistor connected in series. The meter will be connected across the resistor. I adjust the resistor until the reading in the meter read about 5.37V. The value of resistance is 15 ohms. How do I calculate the internal resistance? This is the equation I used

5.37V=10V((Rm*15)/(Rm+15))

I solve for Rm and get 559kOhm. Does this look correct?

2. Sep 11, 2008

### dlgoff

Looks correct to me. Just curious, how did you measure the resistance being 15 ohm? With the same meter?

3. Sep 12, 2008

### alphysicist

I don't see how you are getting 559kOhm from that equation; is there a calculation error?

Also, perhaps I'm visualizing the circuit incorrectly, but could you show how you got that equation?

4. Sep 12, 2008

### bengaltiger14

That 15 ohms was what it took to make the meter reading drop to half of 10V. It is a variable resistor. I tried many ohmage settings and 15Ohms dropped the reading to half of 10V to 5.??.

alphysicist: As for a the miscalculation, you may be right. I calculated that quickly in lab and was unsure if I was even approaching the problem in the correct manner. I will check that but can't do it at the moment.

5. Sep 12, 2008

### alphysicist

What bothered me was that the equation you had in your post was not dimensionally consistent (volts on the left, volts*ohms on the right); I was wondering if there was more to the equation that you had already dealt with that took care of that. Unfortunately I think I am not visualizing your circuit correctly, so there could something that I'm not thinking about.

6. Sep 12, 2008

### dlgoff

So you were using a decade box? There may be tolerances of the resistors to consider when determining the internal resistance of your meter.

7. Sep 12, 2008

### Redbelly98

Staff Emeritus
Are you supposed to calculated the internal resistance of the power supply or of the meter?