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Power question

  • Thread starter Miike012
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  • #1
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Question is in paint doc... question c and d are what I am interested in...

For both questions I used P=IV, this gave me a correct answer for d but not c. For c I got 9/2 the answer is 4.3. Is that close enough? Or am I using the wrong equation?
 

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  • #2
gneill
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Question is in paint doc... question c and d are what I am interested in...

For both questions I used P=IV, this gave me a correct answer for d but not c. For c I got 9/2 the answer is 4.3. Is that close enough? Or am I using the wrong equation?
You should show more of your work. How did you arrive at 9/2 for (c)?
 
  • #3
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P =iv = 0.5a(9.0v)
 
  • #4
gneill
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P =iv = 0.5a(9.0v)
Ah. Well the power supplied by the battery is the power as delivered to external connections at its terminals. The batteries each include an internal resistance, so the voltage at the "9V" battery's terminals is not going to be 9V...
 
  • #5
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is there an equation to find the voltage at the 9V battery terminals?
 
  • #6
gneill
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is there an equation to find the voltage at the 9V battery terminals?
KVL, Ohm's law. What's the potential drop across the internal resistance?
 
  • #7
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Would it be 3?
 
  • #8
NascentOxygen
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Why did you guess 3?
 

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