PhysicsTest said:
As per ohms law the current is generated based on voltage and resistance V=IR. So if 5v and 1k resistor the current 5mA flows in resistor. But if I take a current source, can I pump any amount of current to resistor? So, current drawn is not characteristic of resistor?
I
think you're asking whether Ohm's law holds when using current sources, to which the answer is yes, it does.
Pump an amount of current
i through the resistor, however you achieve that current, and you will find that the voltage across the resistor is equal to
i times
r.
As an exercise, do the following on some paper. Take a voltage source of 5 V and put it in series with a 5 K ohm resistor. This will be your non-ideal current source. Then put both of these in series with a resistive load. Vary the resistance of the load a few times between 50 ohms and 450 ohms. You'll find that the current through the load remains within 5% of 1 mA, which is why this is a current source. In addition, find the voltage drop across both resistors each time you change the load. You'll find that the total voltage drop remains 5V and that the current times the resistance of the load always equals the voltage drop across it.
The thing about current sources that they don't tell you about is how they actually work, which is both good and bad. Good because you don't have to worry about it, but bad because it can confuse you and think that certain things don't seem to apply to them, like Ohm's law. The same is actually true for voltage sources, but the amount of variation in the voltage from the source is generally very small for most circuits, so we don't tend to worry about it very much in early EE courses.
If you're curious about voltage sources, again take some pen and paper and put an ideal voltage source in series with a 0.2 ohm resistor and then connect various loads to it. The voltage source and resistor form a non-ideal voltage source. As you vary the resistance of the load, you'll see that the voltage across the load remains very, very close to 5 V unless you make the resistance very low. As the resistance of the load drops, our non-ideal voltage source drops more and more of the supplied voltage, leaving less for the load.
Again, in all cases Ohm's law holds.
Note that real voltage sources are 'ideal' only across an open, while real current sources are 'ideal' only across a short circuit, which is reflected in the above results as you vary the resistance of the first load closer to zero and the 2nd load closer to infinity. When your load has a finite, non-zero resistance, no source is ideal.