Calculating Internal Resistance in a Series Circuit

  • Thread starter Thread starter bengaltiger14
  • Start date Start date
  • Tags Tags
    Internal
Click For Summary

Homework Help Overview

The discussion revolves around calculating the internal resistance of a meter in a series circuit involving a 10V power supply and a variable resistor. The original poster presents an equation derived from their experimental setup, where the meter reads 5.37V across a 15-ohm resistor.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equation used to calculate internal resistance and question its dimensional consistency. There is also inquiry into how the resistance value was measured and the implications of using a variable resistor.

Discussion Status

Some participants express uncertainty regarding the calculations and the setup of the circuit. There is acknowledgment of potential miscalculations and a request for clarification on the equation used. The conversation indicates that multiple interpretations of the circuit may be present.

Contextual Notes

Participants note the importance of considering tolerances in the resistors used and question whether the internal resistance being calculated refers to the meter or the power supply.

bengaltiger14
Messages
135
Reaction score
0

Homework Statement



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?
 
Physics news on Phys.org
Looks correct to me. Just curious, how did you measure the resistance being 15 ohm? With the same meter?
 
bengaltiger14 said:

Homework Statement



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?

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?
 
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.
 
bengaltiger14 said:
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.

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.
 
bengaltiger14 said:
I tried many ohmage settings and 15Ohms dropped the reading to half of 10V to 5.??.
So you were using a decade box? There may be tolerances of the resistors to consider when determining the internal resistance of your meter.
 
Are you supposed to calculated the internal resistance of the power supply or of the meter?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
941
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 24 ·
Replies
24
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K