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Voltage source and short ckt

  1. Jul 17, 2010 #1
    I don't understand couple of things about voltage and current source.
    When the load impedance is lower than the source impedance of a Voltage source, why does the voltage drop?
    What exactly happens when you draw more current.
    Also, why does ideal current source have a high resistance in parallel with it. Why not just the current source.
  2. jcsd
  3. Jul 17, 2010 #2
    What's the voltage over the RL?

  4. Jul 18, 2010 #3
    Its RL/(RL+Rs).
    I get it now. More voltage is dropped across the internal resistance Rs than the load resistance. I was thinking in terms of internal resistance/load resistance. But never arrived at the answer.

    What about the ideal current source. why a resistor in parallel?
  5. Jul 18, 2010 #4
    It's the practical current source that has a resistor in parallel, typically in the high kΩ-range.

    An ideal current source has an infinite source resistance in parallel (practically an open-circuit), which keeps the circuit current constant when the the load resistance changes.
  6. Jul 18, 2010 #5


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    Isn't the ideal current source modeled with as an infinite voltage source with an infinite resistance in series with it?
  7. Jul 19, 2010 #6


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    That wouldn't achieve what you want as the ratio of two indeterminate quantities would be hard to specify.
    Think of an ideal current source as a supply that will give 'just enough' volts to drive the required (the 'constant') value of current, independent of the value of load resistance. A real current source would have a largeish resistor across it which will sink some of this 'constant' current. The higher the load, the bigger proportion of the supplied current will go through the unwanted parallel path (because the supplied voltage will be higher). So a real constant current source will fail when feeding a high resistance load whereas a real voltage source fails for a low resistance load.
  8. Jul 19, 2010 #7


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    It is the series resistor that is much much larger than the range that the load resistor will be that stabilizes the current. The larger the series resistor (which will require a higher voltage source) the less the current will change as the load resistor changes. Take a load with a range between zero and 1000 ohms. Then put it in series with a resistor thousands of times higher and a supply with enough voltage to give the desired current at a zero ohm load. Do the math and see what happens. When we thevenize a circuit we short circuit all voltage sources and open circuit all current sources. There is a reason for this.
  9. Jul 19, 2010 #8
    Forgive me for my dimness, but wouldn't that leave the current capability of this source undefined? How is this model useful? Also, by modeling the ideal current source with a voltage source, aren't you in effect saying that voltage is more fundamental than current?

  10. Jul 19, 2010 #9


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    You can also model a voltage source with a current source with resistors connected to in various ways. An ideal voltage source or current source only exists in our minds. An ideal voltage source is no different in terms of being undefined. It will supply infinite current. A zero ohm resistor across an ideal voltage source causes ohms law to break down. You cannot divide by zero. Concerning current sources, maybe I should have said series resistors and voltage sources that approach infinity. Have you done the math the way I described?
  11. Jul 20, 2010 #10
    That is the Norton-Thevenin equivalence theorem. But the ideal current source has an infinite parallel impedance. From a power viewpoint, a true current source is better than a voltage source with a large series impedance. In other words, if the desired constant current is 1.0 amp, a 1.0 amp ccs with a parallel impedance of 100 ohms will deliver 0.99 amp into a 1.0 ohm load, 0.999 amp into 0.10 ohms, etc. The power lost in the shunt impedance is 0.01W, 0.001W resp.

    With a voltage source & series resistor, the equivalent is 100V/100 ohms. With the same 1.0 ohm load, the current is 0.99 amp, & with 0.10 ohm load, it is 0.999 amp. But the power lost in the series resistor is 98W/99.8W resp.

    A true current source with a high shunt impedance delivers power with lower losses than its voltage counterpart when the load impedance is smaller than that of the source.

    Last edited: Jul 20, 2010
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