perfect voltage source

a 9 volt alkaline battery with series of 3 ohm resistor behaves as perfect voltage source

what does perfect voltage source means
that 3 ohm already present inside battery as internal,then why we have to supply externally

Bhuvanesh123 said:

a 9 volt alkaline battery with series of 3 ohm resistor behaves as perfect voltage source

what does perfect voltage source means
that 3 ohm already present inside battery as internal,then why we have to supply externally

Click to expand...

A "perfect" voltage source has NO INTERNAL RESISTANCE. Its voltage does not change when its load current changes.
A voltage regulator IC or circuit has a high gain internal error amplifier that makes its output resistance extremely low.

But a 9V alkaline battery voltage changes when its load current changes, look at its datasheet to see it. So it is NOT a perfect voltage source, and with a 3 ohm resistor it is worse.

Bhuvanesh123 said:

a 9 volt alkaline battery with series of 3 ohm resistor behaves as perfect voltage source

what does perfect voltage source means
that 3 ohm already present inside battery as internal,then why we have to supply externally

Click to expand...

Maximum current will flow(without voltage drop) when the source resistance is equal to the load resistance.
Here source resistance should be 3 Ohm

smijesh said:

Maximum current will flow(without voltage drop) when the source resistance is equal to the load resistance.
Here source resistance should be 3 Ohm

Click to expand...

No.
When the source resistance is equal to the load resistance then HALF of the maximum current will flow and the load voltage will be HALF the battery voltage, but the load will have the maximum amount of power.

9V battery with 3 ohms internal resistance:
1) No load. I= 0A, Pl= 9V x 0A= 0W. Pl is power in the load. The most voltage at the load.
2) 6 ohms load. I= 9V/9 ohms= 1A. Pl= 1A squared x 6 ohms= 6W.
3) 3 ohms load. I= 9V/6 ohms= 1.5A. Pl= 1.5A squared x 3 ohms= 6.75W. The most power in the load.
4) 1.5 ohms load. I= 9V/4.5 ohms= 2A. Pl= 2A squared x 1.5 ohms= 6W.
5) Shorted load. I= 9V/3 ohms= 3A. Pl= 0V x 3A= 0W. The most current in the load.