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EMF and Terminal Voltage

  • Thread starter BDR
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  • #1
BDR
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Hello everyone!

I have a question and any help would be appreciated!

(a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A?

(b) Can the emf of the voltage source be found with the information supplied?
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
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BDR said:
Hello everyone!

I have a question and any help would be appreciated!

(a) What is the internal resistance of a voltage source if its terminal voltage drops by 2.00 V when the current supplied increases by 5.00 A?

(b) Can the emf of the voltage source be found with the information supplied?
What do you know about the relationship between voltage current and resistance? Show us some attempt to solve this problem, and then you can get help.

https://www.physicsforums.com/showthread.php?t=4825
 
  • #3
BDR
13
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I know that the internal resistance of a voltage source affects the output voltage when a current flows. I was thinking that I would use the equation V = E - Ir...but I seem to get a different answer each time I use it. I also know that V = IR.

Im my view of this problem I do not have enough information to work the problem with the above equations. Am Ii close?
 
  • #4
OlderDan
Science Advisor
Homework Helper
3,021
2
BDR said:
I know that the internal resistance of a voltage source affects the output voltage when a current flows. I was thinking that I would use the equation V = E - Ir...but I seem to get a different answer each time I use it. I also know that V = IR.

Im my view of this problem I do not have enough information to work the problem with the above equations. Am Ii close?
From V = E - Ir you know what a graph of V vs I should look like. E is the emf (unknown, but constant) and r is the internal resistance (also an unknown constant). The information tells you how much V changes for a change in I. Relate that information to the graph of V vs I and you will have the answer to part a). If you think about trying to draw the graph using the information given, you will also know the answer to part b).
 
  • #5
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BDR said:
I know that the internal resistance of a voltage source affects the output voltage when a current flows. I was thinking that I would use the equation V = E - Ir...but I seem to get a different answer each time I use it. I also know that V = IR.

Im my view of this problem I do not have enough information to work the problem with the above equations. Am Ii close?
also write what you know in equation form:

[tex] V_{1} \ = \ E - I_{1}r [/tex]
[tex] V_{2} \ = \ E - I_{2}r [/tex]

[tex] V_{2} - V_{1} \ = \ (E - I_{2}r) - (E - I_{1}r) [/tex]

[tex] V_{2} \, - \, V_{1} \ = \ -2 \ \mbox{Volts \ \ WHEN} \ \ \ I_{2} - I_{1} \ = \ \mbox{+5 amps} [/tex]

can you now solve for internal resistance r?
 
  • #6
BDR
13
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I'm sorry, but I guess I am not understanding about how I get the information from the graph. Since E and r are unknown constants; does that mean that I could leave them out of the equation and just have V = I? Or am i making it harder than it really is?

Is this close? 2V = E - (5 A)r
 
  • #7
88
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BDR said:
I'm sorry, but I guess I am not understanding about how I get the information from the graph. Since E and r are unknown constants; does that mean that I could leave them out of the equation and just have V = I? Or am i making it harder than it really is?

Is this close? 2V = E - (5 A)r
you didn't simplify the following equation correctly:
[tex] V_{2} - V_{1} \ = \ (E - I_{2}r) - (E - I_{1}r) [/tex]
what happens to E?
 
  • #8
BDR
13
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The E's would cancel each other out? But it seems to me there are more variables than numbers for this problem.
 
  • #9
88
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BDR said:
The E's would cancel each other out? But it seems to me there are more variables than numbers for this problem.
write the full (simplified) equation:
[tex] V_{2} - V_{1} \ = \ (E - I_{2}r) - (E - I_{1}r) \ = \ I_{1}r \, - \, I_{2}r \ = \ (I_{1} \, - \, I_{2})r[/tex]
then use what you know:
[tex] V_{2} \, - \, V_{1} \ = \ -2 \ \mbox{Volts \ \ WHEN} \ \ \ I_{2} \, - \, I_{1} \ = \ \mbox{+5 amps} [/tex]
 
  • #10
BDR
13
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Lets see if i understand: -2 V = 5r......r = -2/5??
 
  • #11
88
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BDR said:
Lets see if i understand: -2 V = 5r......r = -2/5??
almost!
what is the sign of (I1 - I2) in the following equation?
[tex] V_{2} - V_{1} \ = \ (E - I_{2}r) - (E - I_{1}r) \ = \ I_{1}r \, - \, I_{2}r \ = \ (I_{1} \, - \, I_{2})r[/tex]
look carefully. it's a little tricky. remember:
[tex] V_{2} \, - \, V_{1} \ = \ -2 \ \mbox{Volts \ \ WHEN} \ \ \ \color{red}I_{2} \, - \, I_{1} \ = \ \mbox{+5 amps} [/tex]
 
  • #12
BDR
13
0
Is this what you mean.....I1 - I2 would be negative, causing a +2/5...right?
 
  • #13
88
0
BDR said:
Is this what you mean.....I1 - I2 would be negative, causing a +2/5...right?
congratulations!
r = 2/5 ohms
 
  • #14
BDR
13
0
Thanks for the help, its greatly appreciated!!
 

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