Voltage drop in electrical circuits

Question

when I set the DC voltage source of any circuit to a particular level, and then add some loads to the circuit and then re-measure the total voltage of the circuit, it becomes lower than the source's voltage..............is there any reason for that (other than the wires' resistance) ??????

Answer 1

It is some combination of resistance of wires, and the source's internal impedance.  So a batter for example acts like an ideal voltage source connected through a resistor -- so whenever you draw current, the voltage goes down.  

If you are using a regulated voltage source, then it should try to compensate for this -- but if you exceed the current capability then it will either droop, or shut off.

Answer 2

If your source voltage drops under load, then you are seeing the internal resistance of the power supply.  The supply is modeled as a perfect supply in series with a resistance, inside a box.  You only see the result after the internal resistance.  This model works for bench supplies and batteries.  If you ever need to know the value of the hidden internal source resistance, it is the open circuit (no load) voltage divided by the short circuit current.  

Answer 3

It may be due to the internal resistance of the source.