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Find Internal Resistance From 2 Circuits w/ The Same Battery

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Homework Statement


A battery with an emf of 60 V and an internal resistance (r) is connected to a 10 ohm external resistance. The power lost inside the battery is 50w. The same battery is then connected to a 4.0 ohm resistance. The power loss is 200w. What is the internal resistance of the battery?
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Homework Equations



V = emf - Irint

P= VI

V= IR

P=I^2R

P= V^2/R

The Attempt at a Solution


P= V^2/R
V= √ PR
V= √50 x 10
V= 22.36 V

I= V/R
= 22.36/ 10
= 2.23 A
Rint= emf - V/ I
Rint= 60 - 22.3/ 2.23
= 16.9 ohms

This is obviously wrong and I didn't even incorporate the second circuit yet. I'm pretty sure I could finish this question if I knew how to start. Any help is appreciated :)
 
Last edited:

Answers and Replies

  • #2
Orodruin
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Please show your work. Do not just describe it with words.
 
  • #3
CWatters
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Your calculation of V is incorrect. 50w is the power dissipated in the battery yet you used the value of the external resistor.

Note there is a difference between "power delivered" by the battery and "power dissipated" by the battery.
 

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