A generator would need to produce 'x' Volts in order to provide 'y' Amps? Am I understanding this correctly?

Let's say I have a device that requires 190A to work.

My generator has an internal resistance of 2 Ω. The device has a resistance of 0.3 Ω. This would mean that the load on the generator is now 2.3 Ω. Using Ohm's Law IxR=E, the generator would need to produce 437V (not accounting for the voltage drop). I this correct?

And if I add another identical device, then the load is now 2.6 Ω which would mean the generator would now need to produce 494V.

The voltage drop would be 190A x 2 Ω = 380V, so the actual voltage output of the generator would be 57V with just one device or 114V with two devices.

Is any of this correct or am I just way off? Thanks!

• Where did the 2 Ω measurement come from? Is this a real generator with a specification sheet or did you just pick the number out of your head? Jul 16 '20 at 18:33
• "add another identical device" in parallel? Or in series? In normal application "adding on another load" would usually be done in parallel, but the way you just added the new load resistance onto all the other resistances is what you do for series. Jul 16 '20 at 18:52
• @user3439024 If the other device is in parallel then you have two 0.3 Ohm loads in parallel with each other which results in an equivalent 0.15 Ohm load. Then this equivalent load is in series with 2 Ohms so the total resistance seen by the forward BEMF in the generator is 2.15 Ohms, not 2.6 Ohms. It would also mean your generator now needs to supply 380A since each load needs 190A and the current through the first load is not the same current running through the other load as it would be if they were in series. Jul 16 '20 at 19:04
• If they were in series then the current running through the first load is the same as that running through the second load but now you are pushing through double the external resistance which is 0.6 Ohms. Jul 16 '20 at 19:06
• "2 Ω measurement was just arbitrary." I'd say it's too high by a factor of 20 or so. That's going to make your efficiency calculations very poor and unrealistic. Jul 16 '20 at 19:08