American engineer Norton explained his theorem in 1926 after 43 years publishing Thevenin’s theorem. Norton’s theorem is similar to Thevenin’s theorem.
Nortons theorem states that “a linear two terminal circuit can be replaced by an equivalent current source IN in parallel with a resistance RN where RN is the input or equivalent resistance at the terminals and IN is the short circuit current through the terminals when all the independent sources are turned off”.
Figure 1(a) circuit is a linear two terminal circuit. According to Norton’s theorem we replace this circuit by a current source and a equivalent resistance that is figure 1(b).
From source transformation The Thevenin’s and Norton’s resistances are equal
RN = RTh
As two terminal linear circuit is replaced to Norton’s equivalent circuit so they are equal. To find Norton’s current IN from a to b terminal we have to short circuit and that it becomes figure 2.
Norton’s current IN is equal to short circuit current, IN = isc . Independent sources are turned off here as Thevenin’s theorem. As we know RN = RTh and
IN = VTh / RTh
Norton theorem and Thevenin’s theorem are related to source transformation. So source transformation often known as Thevenin-Norton transformation.
As we know VTh = Voc , IN = isc
RTh = Voc / isc = RN .
To find Thevenin’s or Norton’s equivalent short circuit and open circuit tests are sufficient if one side of the circuit contains at least one independent source.
Norton’s theorem problem:
Find Norton’s equivalent circuit from following circuit and also find IN
Norton theorem problem Answer:
we have to find RN as the same as we find RTh in Thevenin’s equivalent circuit. First we turn off all independent source equal to zero as we get figure 4(a).
Here 8, 4, 8Ω are in series and their equivalent resistance is in parallel with 5Ω resistance. Thus we get
RN = 5 ││ ( 8 + 4 + 8 ) = 20 = (20×5) / 25 = 4Ω
We have to short circuit a and b terminals to find IN then the circuit becomes figure 4(b).
Ignoring 5Ω resistance cause it is short circuited we apply mesh analysis and obtain,
i1 = 2A, 20i2 – 4i1 – 12 = 0
From above equation we get, i2 = 1A = isc = IN
We can also determine IN from the value of VTh / RTh . Making open circuit terminals a and b we get VTh from figure 4(c).
Applying mesh analysis,
i3 = 2A,
25i4 – 4i3 – 12 = 0
or i4 = 0.8A
and voc = VTh = 5i4 = 4v
hence IN = VTh / RTh = 4/4 = 1 A
as we got it previously. This is the confirmation of RTh = Voc / isc = 4/1 = 4Ω = RN.
Thus we get Norton’s equivalent circuit as shown in figure 5.
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