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Superposition theorem, constant current sources' internal resistance

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Hi,

I've been told that when using the Superposition theorem to analyse a circuit, I should replace all constant current sources other than the one I'm analysing with an open-circuit. But I need to retain the internal resistance of the current source in the circuit. Where does that resistance go, if the current source is replaced with an open?

Richard
 
Hello,

You do not need to retain the internal resistance of a current source when it is 'killed'. The idea is to eliminate the effects of that current source completely while we look at the effects of the other sources. If you try to keep some part of that current source in the circuit while you look at the effects of the other sources then you havent killed it properly. To kill it properly it gets open circuited.

One way to look at this is to think of the current source as an infinite impedance as seen by the OTHER sources. For example, when we put a voltage source in parallel with a resistor and increase that voltage by some increment we see a corresponding change in current through the resistor, but when we put a voltage source in parallel with a current source and increase the voltage by some increment we do not see any change in current, and that's the same thing that happens when we put a voltage source in parallel with an infinite resistance and increase the voltage by some increment.
 
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Okay, so I replace all voltage sources with a resistor that represents that voltage source's internal resistance, and I replace all current sources with an open?
 
Hello again,

If you are talking about a theoretical voltage source you simply short it out. If it is a practical voltage source then it has some series resistance, and you could just draw that in the circuit before shorting out the voltage source.
Yes, simply replace all the current sources with an open circuit.
 
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