EMF and Circuits

  • #1

Homework Statement



In the circuit (http://i.imgur.com/II38M.jpg),a variable resistor R is connected across the terminals of a battery, and an ideal ammeter and ideal voltmeter are also connected. In the figure, E is the internal emf of the battery, and r is its internal resistance. All the connecting cables have no appreciable resistance. The ammeter reads 3.10 A when the voltmeter reads 5.08 V, and the voltmeter reads 15.0 V when the ammeter reads 0.00 V.

Homework Equations



Vab = E - IR


The Attempt at a Solution



I thought since an ideal voltmeter has infinite resistance, so there would be no current through the resistor. I thought it should just be 15V then, but I feel like I am missing something. Any guidance would be greatly appreciated, thanks.
 

Answers and Replies

  • #2
gneill
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What, precisely, were you asked to determine? It looks to me like you're in a position to determine both E and r.
 
  • #3
I am sorry, don't know how I forgot the most important part. It asks when R = 6.00 ohms, what is the reading of the voltmeter
 
  • #4
gneill
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If you knew the values of E and r, then when R is made 6.00 Ω you can calculate the voltage across it (it'll be a simple voltage divider circuit). So, can you think of how you might determine E and r from the given data?
 
  • #5
Well, V = E - Ir, would you solve for r to be 5.08V = 15V - 3.1A*r? And I am sorry, but I don't quit get what you mean by a voltage divider, they split evenly?
 
  • #6
gneill
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Well, V = E - Ir, would you solve for r to be 5.08V = 15V - 3.1A*r? And I am sorry, but I don't quit get what you mean by a voltage divider, they split evenly?
Yes. You can find r that way. I see that you've also decided that E = 15V (the open circuit voltage across the battery). That's correct.

A voltage divider using resistors is a series connection of two resistors, with the "output" voltage taken across one of the resistors. So, if you had resistors R1 and R2 connected to a supply voltage V, and the output was the potential across R2, then

Vout = V*R2/(R1 + R2)

This follows from the fact that the current through the series connected resistors is V/(R1 + R2), so the voltage across R2 must be R2* V/(R1 + R2).
 
  • #7
Awesome, thank you so much for all the help. That makes perfect sense, sorry I was confused. You really helped out a lot, have a great day
 

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