The generators can be grouped in two ways, serial and
parallel:
1. Grouping in series mode.
To group in serial mode a couple of generators, we have to
mount the negative terminal of the first generator with the positive terminal
at the second generator and the action continues in this manner from all the
generator that we have.
Let's consider three generators E1,E2 and E3 with the internal resistor r1,r2 and r3 all connected in series wich carry a resistive consummer noted with R.
By applying the Kirchhoff law, the second one, on our circuit, we will obtain: E1+E2+E3=IR+Ir1+Ir2+Ir3.
which result:
Compared with the Ohm's law into a closed circuit:
trough series mode connection of three generators, we consider:
2. Grouping in parallel mode.
For grouping in parallel mode, a couple of generators, we have to connect all the positive terminals of the generators together and all the negatives terminals of the generators as well.
We consider three identical generators with e.m.f. E and internal resistance r, grouped in parallel and fed a consumer with a resistance R. Applying Kirchhoff's laws on the circuit we will obtain:
But: I1=I2=I3 so I=3I1
which result:
Compared with the Ohm's law into a closed circuit:
trough series mode connection of three generators, we consider:
- electromotive force is equal with sum of Electromotive force(e.m.f) of our generators: E=E1+E2+E3.
- internal resistance is the sum of the generators resistances: r = r1 + r2 + r3.
2. Grouping in parallel mode.
For grouping in parallel mode, a couple of generators, we have to connect all the positive terminals of the generators together and all the negatives terminals of the generators as well.
We consider three identical generators with e.m.f. E and internal resistance r, grouped in parallel and fed a consumer with a resistance R. Applying Kirchhoff's laws on the circuit we will obtain:
I=I1+I2+I3.
E=I1r+IR.
But: I1=I2=I3 so I=3I1
We have:
Results the fact that e.m.f is E but the internal resistance become r/3.
