Let a linear network consisting of one or more independent sources and linear elements be feeding power to a load. Norton’s theorem says that the linear network consisting of one or more independent sources and linear elements can be represented by a current source and an equivalent resistance in parallel with the current source. To find the value of the current source, replace the load by a short-circuit. The value of the current source is the current that flows through the short-circuit. The resistance is the resistance of the network as viewed from the load terminals, with the independent sources removed. This resistance is known as Thevenin’s resistance. The current source is known as the Norton’s current source.

Let a circuit be represented by a box, as shown in Figure 28. Its output characteristic is also displayed. As the load resistor is varied, the load current varies. The load current is bounded between two limits, zero and ISC, and the load voltage is bounded between two limits, VTh and zero volts. When the load resistor is infinite, it is an open circuit. In this case, the load voltage is at its highest, specified as VTh and the load current is zero. This is the point at which the output characteristic intersects with the Y axis. When the load resistor is of zero value, there is a short circuit across the output terminals of the circuit and in this instance, the load current is maximum, specified as ISCand the load voltage is zero. It is the point at which the output characteristic intersects with the X axis.

The circuit in Fig. 29 has an
output current of ISC, when the load voltage is
zero. Hence the model of the circuit can have a current source of
ISCamperes. When the output
terminals are open circuited, the load current is zero and it can be stated
that the source current passes through the internal resistance, leading to a
terminal voltage of VTh. This means that the internal
resistance of the circuit, called as RTh, has a value, as shown by the
equation displayed. Hence the circuit consists of a current source of value
ISCamperes and a resistor of value
equal to RTh. This resistor is the
resistance of the circuit, as viewed from the load terminals, after removal
of the independent sources in the circuit. A voltage source is replaced by a
short-circuit, and a current source is replaced by an open circuit. The
internal resistance of each of these sources is left in the circuit, as it
is, where it is. A non-ideal voltage source consists of a voltage source with
a source resistance, connected in series with the source. A non-ideal current
source consists of a current source with a source resistance, connected
across the source.

The circuit in Fig. 30, which has already been presented earlier, is shown here for illustrating Nortons theorem. The short circuit current and the load current can be obtained, as shown by the simplified circuits. We can describe the operation of the circuitby equations described below.

Equation (54) presents the resistance seen by
the source. Equation (55) shows how the current drawn from the source can be
obtained. The source current flows into resistors, R2 and
R3, connected in
parallel. Equation (56) expresses the current through resistor
R3.

For the circuit in Fig. 30, let
R3 be considered the load resistor. Then the load current
can be expressed by equation (57). This equation is obtained by replacing the
source current in equation (56) in terms of the source voltage and the
resistances. From equation (57), expressions for the short circuit current,
and Thevenin’s resistance can be obtained, and equation (58) presents
the expressions . It is seen that the load current can be expressed in terms
of Norton/s current and Thevenin’s resistance.

Based on the example, Norton’s theorem can be stated as follows.Let a network with one or more sources supply power to load resistor as shown in Fig. 31. Norton’s theorem states that the network can be replaced by a single equivalent current source, marked as ISCand a resistor marked as RTh, as shown in Fig. 31.

The circuit in Fig. 32 can
be used to prove Norton’s theorem.
Equation (1) expresses current source IY connected to the
load terminals, as a function of voltage VYacross the source and some
coefficients.
Coefficient k1 reflects conductance of the circuit as seen
by external current source IY, and coefficientk2reflects the contribution to
terminal voltage by internal sources and components of the circuit. It is
valid to do so, since we are dealing with a linear circuit. Each independent
internal source within the circuit contributes its part to terminal voltage
and coefficientk2is the algebraic sum of
contributions of internal sources. There is a negative sign in front of
k1in equation (1). Let voltage
VYbe a a positive value. Then as
current IYincreases, voltage
VYtends to decrease. By assigning
a negative sign in front of k1, we get apostivie value for
k1. Adjust external current
source such that voltage VYbecomes zero. As shown by
equation (2), coefficientk2is Norton’s current. To
determine Thevenin’s resistance, set external source current to zero.
Then in this case, the voltage across the load terminals is the open circuit
voltage or Thevenin’s voltage. Coefficient
k1is the reciprocal of
Thevenin’s resistance. This concludes the proof of Norton’s
theorem.
The application of Norton’s theorem is similar to the application of Thevenin’s theorem. It can be illustrated with the help of a sketch.

Steps involved in obtaining
Nortons equivalent circuit are outlined. For the generic block diagram in
Fig. 33, Obtain the short circuit current first, as described.

Thevenin’s resistance is obtained as described above. These steps have been described earlier, for getting Thevenin’s resistance for Thevenin’s theorem.
Norton’s Theorem and
Thevenin’s Theorem are equivalent, as we will see now.
Their equivalence leads to
source transformation.

The relationship between the Thevenin’s theorem, and, Norton’s theorem can be established using the equivalent circuits in Fig. 34.
From Fig. 34, the load current, obtained using Thevenin’s equivalent circuit, is expressed by equation (59). The load current obtained using Norton’s equivalent circuit is expressed by equation (60). Both the expressions in equations, (59) and (60) yield the same value, when equation (61) is valid. Equation (61) describes the relationship between the Thevenin’s theorem, and the Norton’s theorem. The Thevenin’s voltage equals the product of Norton’s current and the Thevenin’s resistance. We can use the equivalence between Thevenin’s equivalent circuit, and Norton’s equivalent circuit to show how source transformation can be brought about.
What is meant by source transformation is that a voltage source, with a resistor in series, can be effectively replaced by a current source, and, a resistor in parallel. The voltage source equals the product of the value of the current source, and, the resistance in series with the voltage source. What is being stated is, that Thevenin’s equivalent circuit can be replaced by the Norton’s equivalent circuit, and, vice versa. The source transformation theorem is a direct consequence of Thevenin’s and Norton’s theorems. Figure 35 illustrates the process of source transformation. Source transformation can be used for simplification of some circuits. In some instances, we can derive Thevenin’s and Norton’s equivalent circuits using source transformation , as illustrated by the example presented below.
A problem is presented now by the circuit in Fig. 36. The task is to obtain the Thevenin’s equivalent circuit, and, the Norton’s equivalent circuit by using source transformation. The first step is displayed next.
The current source of 2 Amperes, and, 3 W resistor can be replaced by a voltage source of 6 Volts, with the resistor connected in series. The resulting circuit is displayed in Fig. 37.
The two voltage sources can be combined into a single voltage source, with the resistance in series being the sum of 3 W and 5 W . The resultant circuit is displayed by circuit in Fig. 38.
The circuit in Fig. 39 displays the next step. The voltage source of 16 Volts, and, the 8 W resistor can be replaced by a current source of 2 Amperes, with the 8 W connected across it. The two current sources in parallel can be replaced by a single current source, and, the two resistors in parallel can be replaced by the equivalent parallel value.
The circuits in Figure 40 display the Thevenin’s and the Norton’s equivalent circuits. A few interactive examples are presented in the subsequent pages.
This page has presented Norton’s theorem and source transformation. It is seen that Norton’s theorem and Thevenin’s theorem are linked, and they form the basis for source transformation. A few interactive examples are presented in the subsequent pages.After that, maximum power transfer theorem is described.