Norton’s
Equivalent
circuit
Fig. 1(a)
Fig. 1(b)
According to Norton’s theorem, the linear circuit in Fig. 1(a) can be
replaced by that in Fig. 1(b).
 Now, the question is how to get RN
and IN?
 We have to find RN in the same
way we find RTh.
 In fact, the Thevenin and Norton
resistances are equal; i.e.,
RN = RTh
Fig. 2
 To find the Norton current IN, determine the short-circuit current
flowing from terminal a to b in both circuits in Fig.1.
 The short-circuit current in Fig.1(b) is IN. This must be the same
short-circuit current from terminal a to b in Fig.1(a), since the two
circuits are equivalent. Thus,
IN = isc
 shown in Fig.2. Dependent and independent sources are treated
the same way as in Thevenin’s theorem.
Fig. 2(b)
 Observe the close relationship between Norton’s and Thevenin’s
theorems:
RN = RTh, and IN = VTh/RTh
Since VTh, IN, and RTh are related, to determine the Thevenin or
Norton equivalent circuit requires that we find:
• The open-circuit voltage voc across terminals a and b.
• The short-circuit current isc at terminals a and b.
• The equivalent or input resistance Rin at terminals a and b when all
independent sources are turned off.
We can calculate any two of the three using the method that takes
the least effort and use them to get the third using Ohm’s law. Also,
since
RTh
VTh = voc
IN = isc
= voc / isc = RN
the open-circuit and short-circuit tests are sufficient to find any
Thevenin or Norton equivalent.
Example: Find the Norton equivalent circuit of the circuit in the
given Figure.
Solution:
Fnd RN in the same way we find RTh in
the Thevenin equivalent circuit. Set
the independent sources equal to
zero. This leads to the circuit as in
Fig. a:
Now, find RN. Thus,
RN = 5 // (8 + 4 + 8)= 5 // 20 =
20 × 5 /25 = 4 ohm
To find IN, short-circuit terminals a and
b, as shown in Fig. below:
We ignore the 5-resistor because it has been short-circuited.
Applying mesh analysis, we obtain
i1 = 2A, 20i2 – 4i1- 12 = 0
From these equations, we obtain
i2 = 1A = isc = IN
Alternatively, we may determine IN from VTh /RTh. We obtain VTh as the
open-circuit voltage across terminals a and b in Fig.(c). Using mesh
analysis, we obtain
i3 = 2A,
25i4 – 4i3 - 12 = 0 ⇒ i4 = 0.8A
and, voc = VTh = 5i4 = 4V
Hence,
IN = VTh /RTh = 4/4 = 1A
This also confirms, RTh = voc /isc = 4/1
= 4 ohm .
Thus, the Norton equivalent circuit is as
shown in Fig. (d).
(d)
Example:
Obtain RN and IN at terminal 1 and 2 of the following circuits.
Solution:
(a)
RN = RTh = 10//40 = 8 ohms.
To find IN, consider the circuit in Fig. (a).
Apply Short circuit & then find Isc.
40Ω
IN = Isc = 20/10 = 2 A
(b)
RN = RTh = 30//60 = 20 ohms
To get IN, consider the circuit in Fig. (b).
IN = Isc = 2 + 30/60 = 2.5 A
Example:
Obtain RN and IN at terminal 1 and 2 of the following circuit.
Solution:
1
For RN , consider the circuit in Fig. (a).
RN = (6 + 6)||4 = 3 ohms
For IN , consider the circuit in Fig. (b).
The 4-ohm resistor is shorted so that
4-A current is equally divided between
the two 6-ohm resistors. Hence, IN =
4/2 = 2 A
2
Example: Find the Norton equivalent looking into terminals a-b of the circuit in
the given Fig.
Solution:
RN = RTh = 20 + 10//40 = 28 ohms
To find IN, consider the circuit below,
At the node,
(40 – vo) / 10 = 3 + (vo / 40) + (vo / 20)
Or, vo = 40 / 7
io = vo / 20 = 2/7, but IN = Isc = io + 3
= 3.286 A
Example: Determine the Norton quivalent at terminals a-b for the given circuit.
To get RTh, consider the circuit in Fig. (a).
From Fig. (a), Io = 1, 6 – 10 – V = 0, or V = -4
RN = RTh = V/1 = -4 ohms
To get VTh, consider the circuit in Fig. (b),
Io = 2, VTh = -10Io + 4Io = -12 V
IN = VTh/RTh = 3A