Tricky V I EMF Graph Need help please

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Offline Manager123

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Tricky V I EMF Graph Need help please
« on: 07/06/2013 09:34:15 »

Im really confused with this question, newbielink: [nonactive]

the way i did it was like this  newbielink: [nonactive]

However the voltmeter is over the battery, so does this mean that its the opposite and a positive gradient ???


Offline evan_au

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Re: Tricky V I EMF Graph Need help please
« Reply #1 on: 08/06/2013 02:12:40 »
The answer just gives you one quadrant to graph the results, so it presumes that the voltmeter V and current meter A are connected to give positive readings.

The graph you give looks right:
  • When there is zero current, the voltage measured is ε Volts
  • When the variable resistor=0, the voltage will be zero, and the current will be at a maximum = ε/r Amps.
  • The equation of the line is V=ε-rI
  • But the gradient is "-r", not "r", giving a positive resistance

As a "sanity check", this looks right because, with realistic passive components:
  • Resistors are ≥ 0 Ohms
  • The battery output power is finite:
    • If you put too much load on it, the power output (VxA) drops to zero because V=0
    • If you don't put any load on it, the power output (VxA) is zero because I=0
    • There is some maximum output power in the middle, where the variable resistor is the same as r
  • The idea of having a circuit where the slope of the line is positive means that as you increase the load, the output voltage also increases
    • So if you reduced the load to zero, the battery output power would be infinite!
    • This is quite unlike ordinary batteries!
PS: This assumes "ideal" measuring instruments: A Voltmeter with infinite impedance, and an Ammeter with zero impedance.


Offline syhprum

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Re: Tricky V I EMF Graph Need help please
« Reply #2 on: 08/06/2013 11:11:54 »
Ohms law is of course a poor representation of how things behave in the real world if Manager123 measured the ratio of current to voltage in his lamps he would find that it varies by a factor of ten to one depending upon how much current is flowing.