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Power Supplied by the Leftmost element

  • Thread starter Mosaness
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1. Determine the power supplied by the leftmost element in the circuit



Homework Equations



Power = Voltage ( Current)



The Attempt at a Solution



As power = voltage (current), that is the equation that will be used. The leftmost element shows a current of 2 AMPS entering through the negative node and a voltage of 2V.

Therefore the power ought to be :

P = (-2A)(2V) = -4W

I just wanted to know if I did this correctly as we do not have a solutions manual for the book we are using.
 

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  • #2
CWatters
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The question asked how much power it was supplying (eg to the rest of the circuit). That would be +4W.
 
  • #3
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The question asked how much power it was supplying (eg to the rest of the circuit). That would be +4W.
I am still slightly confused as to why that would be so? Why 4 W. Why not -4W? We were told in class that it could be a negative power as well.
 
  • #4
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Is it because both the current and the voltage are entering trough the negative node?
 
  • #5
NascentOxygen
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I am still slightly confused as to why that would be so? Why 4 W. Why not -4W? We were told in class that it could be a negative power as well.
The element on the left has a terminal marked +. Its current is shown leaving that terminal, flowing through a circuit external to that element, to return to its negative terminal. That represents power output from that element, the element is doing work, delivering energy to the external circuit. It's precisely how a battery-powered circuit is drawn, showing current emerging from the battery's + terminal.

Were the current to be drawn reversed, so that it enters the element at its + terminal, that represents work being done on that element, the external circuit is forcing current into the element's + terminal and doing work on it. That's a typical arrangement when a secondary battery is being recharged. So power coming from the element would be negative (because power is going to it, not from it).
 
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  • #6
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The element on the left has a terminal marked +. Its current is shown leaving that terminal, flowing through a circuit external to that element, to return to its negative terminal. That represents power output from that element, the element is doing work, delivering energy to the external circuit. It's precisely how a battery-powered circuit is drawn, showing current emerging from the battery's + terminal.

Were to current to be drawn reversed, so that it enters the element at its + terminal, that represents work being done on that element, the external circuit is forcing current into the element's + terminal and doing work on it. That's would be a typical arrangement when a secondary battery is being recharged. So power coming from the element would be negative (because power is going to it, not from it).
And the Voltage is also shown leaving the positive terminal, which gives us a positive power correct?
 
  • #7
NascentOxygen
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And the Voltage is also shown leaving the positive terminal, which gives us a positive power correct?
We don't usually speak of a voltage moving or leaving. Current being comprised of a flow of charge carriers can be pictured as moving or leaving. Voltage is present at a point, it exists at a point (and is measured relative to the voltage at another point).

But yes, the voltage at the + terminal is positive (with reference to the other terminal), and current at the + terminal is in the positive direction for doing positive work. With voltage and current both being positive, so is power! :smile:
 
  • #8
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We don't usually speak of a voltage moving or leaving. Current being comprised of a flow of charge carriers can be pictured as moving or leaving. Voltage is present at a point, it exists at a point (and is measured relative to the voltage at another point).

But yes, the voltage at the + terminal is positive (with reference to the other terminal), and current at the + terminal is in the positive direction for doing positive work. With voltage and current both being positive, so is power! :smile:

Thanks! I am still fairly new at Circuit Theory, so I am kind of struggling! But hey, practice makes perfect!
 

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