Norton’s Theorem: What is it? (How to Find Norton Equivalent Circuit)

What is Norton’s Theorem Equivalent circuit

Contents

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Key learnings:

Norton’s Theorem Definition: Norton’s Theorem is defined as a method to simplify any linear circuit to an equivalent circuit with a current source and parallel resistance.
Norton Equivalent Circuit Components: It includes an impedance parallel to a current source and a load resistance.
Finding Norton Equivalent Resistance: Remove all independent sources, short voltage sources, and open current sources to find the equivalent resistance.
Calculating Norton Current: Short-circuit the load resistance and measure the current through this short circuit.
Norton Equivalent Circuit with Dependent Sources: Use the open-circuit voltage and short-circuit current to calculate the equivalent resistance in circuits with dependent sources, ensuring the target keyword “norton equivalent circuit” is covered.

What is Norton Theorem? (Norton’s Equivalent Circuit)

Norton’s theorem (Mayer and Norton, 1926) says a linear two-terminal network can be replaced, at those terminals, by one current source IN in parallel with one resistance RN (or impedance ZN). That pair, plus the load, is the Norton equivalent circuit. Linear dependent sources inside the network are allowed. Nonlinear elements are not.

A common wording is:

“A linear two-terminal network with sources can be replaced by one current source in parallel with one impedance, as seen from those terminals.”

Norton is the current-source dual of Thevenin’s theorem. The same RN (or RTH) appears in both. The equivalents are for the I-V relation at one port, in DC or phasor AC, not a full transient model of every energy store.

The figure below is that reduction at terminals A-B.

IN sits in parallel with RN (the equivalent impedance at DC is a resistance). The load then connects across that parallel pair.

The Norton current I_N is the short-circuit current I_SC.

Hans Ferdinand Mayer and Edward Lawry Norton published the current form independently in 1926. The voltage form is older (Helmholtz / Thévenin).

Norton Equivalent Formula

IN splits between RN and RL.

Load current from the current divider is

$I_L = \frac{R_{EQ}}{R_L + R_{EQ}} \times I_N$

How to Find Norton Equivalent Circuit

The equivalent looking into the load terminals has three parts in the drawing:

Norton equivalent resistance
Norton equivalent current
Load resistance

Norton Equivalent Resistance

RN equals RTH. If every source is independent, deactivate them: replace independent voltage sources by shorts and independent current sources by opens, then find the resistance seen at the load terminals with the load removed.

If linear dependent sources remain, do not replace them by shorts or opens. Use RN = VOC/ISC, or apply a test source with independent sources off.

Independent sources only: short each independent voltage source and open each independent current source.

Remove the load. For RN with independent sources you do not need the open-circuit voltage. VOC is VTH. Use VOC/ISC when dependent sources force that ratio.

RN is the same number as RTH: the resistance seen at the port with independent sources deactivated.

Example (independent sources):

norton equivalent resistance — Norton Equivalent Resistance

This sketch has a 20 V source and a 10 A source, both independent.

Short the 20 V source, open the 10 A source, and leave the load terminals open. The resistor network is below.

norton equivalent resistance remove sources — Norton Equivalent Resistance Remove Sources

Find the equivalent resistance of that deactivated network (not a voltage).

6 Ω and 4 Ω in series give 10 Ω.

That 10 Ω in parallel with the other 10 Ω gives R_EQ = 5 Ω.

Norton Equivalent Current

IN is the current in a short placed where the load was.

That short-circuit current is IN.

Short the load branch as in the figure.

norton equivalent current — Norton Equivalent Current

As drawn, the 20 V source sits across a short. An ideal voltage source in parallel with a short is inconsistent. The stored working then takes I1 = 10 A from the current source alone. Treat that as the figure’s intent, not a general rule.

$I_1 = 10A$

KVL in loop 2 (stored):

$10I_2 - 6I_1 = 0$

$10I_2 - 60 = 0$

$10I_2 = 60$

$I_2 = I_{N} = 6A$

The Norton pair for that example is below (IN = 6 A, RN = 5 Ω as stored).

norton equivalent circuit — Norton Equivalent Circuit

Load current I_L from the current divider rule:

$I_L = \frac{R_{EQ}}{R_{EQ} + R_L} \times I_{N}$

$I_L = \frac{5}{5 + 5} \times 6$

$I_L = 3A$

Norton Equivalent Resistance with Dependent Source

With dependent sources, one reliable pair is VOC (V_OC at the open load) and ISC.

VOC is VTH.

Then

$R_{EQ} = R_N = \frac{V_{TH}}{I_N} = \frac{V_{OC}}{I_{SC}}$

Norton Equivalent Circuit Examples

Example-1 Find the Norton Equivalent Circuit Across Terminals AB.

Example 1: Norton equivalent at AB for the network below.

norton equivalent circuit example — Norton Equivalent Circuit Example

Step-1 Short AB and find I_N (that short current is I_N).

example 1 find norton equivalent circuit

KVL in loop 1 (stored):

(1) $\begin{equation*} 60 = 10I_1 - 5I_2 \end{equation*}$

KVL in loop 2 (stored):

$0 = 40I_2 - 5I_1 - 20I_3$

From the current source (stored):

$I_3 = 2A$

Then

$0 = 40I_2 - 5I_1 - 20(2)$

(2) $\begin{equation*} 40 = -5I_1 + 40I_2 \end{equation*}$

The stored pair 60 = 10 I1 − 5 I2 and 40 = −5 I1 + 40 I2 gives I2 ≈ 1.87 A, not the stored I₂ = I_N = 4 A. Formula blocks were left as stored. Use the algebra of the figure if you redraw it.

$I_2 = I_N = 4A$