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DC Network Analysis

# Norton's Theorem

Norton's theorem provides a second method of reducing complex circuitry to a simple equivalent circuit.

Norton's theorem is especially useful in analyzing circuits where only one
particular resistor in the circuit (called the **load**) is subject to
change. Example:

Norton's theorem states: *Any linear electrical circuit can be replaced
at terminals A and B by a simple parallel circuit consisting of an equivalent
current source* *I*_{no} * and an equivalent resistance*
*R*_{no}.

The equivalent parallel circuit will provide the same current through a load
as would the original circuit. The equivalent current *I*_{no}
is the current that would flow through a short circuit between the two
terminals (A and B) being considered in the original network.
The equivalent resistance *R*_{no} is the resistance "seen"
between the two terminals being considered in the original network when all
voltage sources of the original circuit are replaced by a short circuit (wire)
and all current sources are replaced by an open circuit (break).

The figure below shows the application of Norton's theorem to a simple network.
When Norton's theorem is applied, the current through *R*_{L}
of the figure below (view A) is the same as the current in *R*_{L} of
the figure below (view D).

In Norton's theorem, as in Thevenin's theorem, the equivalency is established for the chosen load terminals. Therefore, the equivalency applies only at the load.

Application of Norton's theorem is further illustrated in the following examples.

**Example 1:**

Find the Norton-equivalent circuit of the circuit shown in the figure below.

Solution:

**Calculating the equivalent current I_{no}**

*I*

_{no}.

Place a wire (short) between the points A and B (figure above), and use Ohm's
law to calculate the currents through resistors *R*_{1} and
*R*_{2}:

The current through the short between the points A and B is the sum of the currents
*I*_{R1} and *I*_{R2}:

**Calculating the equivalent resistance R_{no}**

Replace the voltage sources *V*_{1} and *V*_{2} with
a wire:

*R*

_{no}.

The resistance between the points A and B is equal to *R*_{1} and
*R*_{2} in parallel: 0.8 Ω. This is our equivalent
resistance *R*_{no} in the equivalent circuit:

**Example 2:**

Find the Norton-equivalent circuit of the circuit shown in the figure below.

Solution:

**Calculating the equivalent current I_{no}**

*I*

_{no}.

Place a wire (short) between the points A and B (figure above), and use
Ohm's law to calculate the currents through resistor *R*_{2}:

The current through the short between the points A and B is the sum of the
currents *I*_{R1} and *I*_{R2}:

**Calculating the equivalent resistance R_{no}**

Replace the current source *I*_{1} with a break and the voltage
source *V*_{2} with a wire:

*R*

_{no}.

The resistance between the points A and B is equal to *R*_{2}: 1 Ω.
This is our equivalent resistance *R*_{no} in the equivalent circuit: