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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!

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  • \$\begingroup\$ 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? \$\endgroup\$ – Transistor Jul 16 '20 at 18:33
  • \$\begingroup\$ "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. \$\endgroup\$ – DKNguyen Jul 16 '20 at 18:52
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    \$\begingroup\$ @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. \$\endgroup\$ – DKNguyen Jul 16 '20 at 19:04
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    \$\begingroup\$ 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. \$\endgroup\$ – DKNguyen Jul 16 '20 at 19:06
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    \$\begingroup\$ "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. \$\endgroup\$ – Transistor Jul 16 '20 at 19:08
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437V is accounting for the voltage drop [across the generator resistance]. 437V is what is required to push 190A through 2.3 Ohms, some of which is the load and some of which is the generator's resistance. 57V is the voltage you would need not accounting for the voltage drop since If the generator had no resistance you would only need the generator to produce 57V to get 190A through the 0.3 Ohm load.

By the way, you would not actually measure 437V if you took probes to the generator terminals while it is powering the load. You would measure 57V, because the remaining 380V has already been dropped across the gen's resistance is inside the generator by the time it reaches the output terminals.

In your scenario you're losing way more power and voltage inside the generator than is actually getting the load. This is pretty much what happens when you connect a generator that is too small to a load that is too large.

Normally you choose a generator with an internal resistance much less than your load resistance so that the power (and voltage) delivered to the load is the vast majority and that which is lost inside the generator is the minority. Otherwise, it's a bit self-defeating. The example you gave would probably qualify as shorting out the generator since the load resistance is so small relative to the generator's own resistance.

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Given an AC generator with 2 ohms internal resistance, the answer would be to use a step-down transformer, say, 4.5:1 in voltage, or 20:1 in impedance, to step that 2 ohm impedance down to 0.1 ohm at the secondary.

To generate 190 amperes through 0.3+0.1 = 0.4 ohms, you need 76V at the secondary, or 342V on the primary, and only 42A from the generator. This is considerably more efficient than the direct connection. (I haven't allowed for transformer losses, which may be around 5 to 10%)

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