How Does Current Change with Resistance in Constant Voltage Sources?

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In a constant voltage source, as the resistance of a variable resistor changes from zero to infinity, the current through the resistor varies according to Ohm's Law, I = V/R, where voltage remains constant. For an ideal constant voltage source, the internal resistance is zero, allowing maximum current draw without voltage drop. In contrast, a real constant voltage source has internal resistance, which causes a voltage drop when current flows, limiting the maximum current available. As resistance approaches zero in a real source, the current draw increases until it is limited by the internal resistance. Understanding these differences is crucial for analyzing circuit behavior in electrical engineering.
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Hey guys

I am new to Electrical Eng, i got a question to answer, it says that a variable resistor is connected to a " Constant Voltage Source", how does the current through the resistor changes as its resistance varies from zero to infinity?

I also need with the same question, but with a ideal constant Voltage source and real constant vol source!

Thanks
 
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Do you know Ohms Law? If so, see how the current is related to voltage and resistance. Remember, the voltage is constant.
 
Yeh V = R*I, Since for the first V = Constant, Then I = V/R, I = 1/r, but what is the difference between ideal, real, and voltage constant sources!,

thanks
 
There are many subtle differences, and one big one. What happens to the current draw from an ideal source as the load resistance goes to zero? Would you expect the same behavior from a real source?
 
Can u give me some definitions for ideal and real sources and how the ohm's law works with those! And how resistance affects the current in those sources.

thanks and sorry really new with circuits and all!
 
Real voltage sources have an internal resistance - draw it as a battery with a small resistor in series. Now image you have a 12V battery with an internalal resistance of 1Ohm, as you draw 1Amp from the battery there is a 1V drop across the internal resistor and so you only get 11V out. What's the maximum current you can draw from such a battery?
AN ideal source has a zero Ohm internal resistance.
 
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