In 1926, about 43 years after Thevenin published his theorem, E.L. Norton, an American engineer at Bell Telephone, proposed a similar theorem.

This circuit analysis theorems are classified as:

This circuit analysis theorems are classified as:

Before learning about Norton's theorem, let us read the brief explanatin below :

Norton's theoremstates that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source I_{N}in parallel with a resistor R_{N}, where I_{N}is the short-circuit current through the terminals and R_{N}is the input or equivalent resistance at the terminals when the independent sources are turned off.

Thus, the circuit in Figure.(1a) can be replaced by the one in Figure.(1b).

Figure 1 |

_{N}in the same way we find R

_{Th}. In fact, from what we know about source transformation, the Thevenin and Norton resistances are equal; that is,

(1) |

To find the Norton current I

_{N}, we determine the short-circuit current flowing from terminal a to b in both circuits in Figure.(1). It is evident that the short-circuit current in Figure.(1b) is I_{N}. This must be the same short-circuit current from terminal a to b in Figure.(1a), since the two circuits are equivalent. Thus,(2) |

shown in Figure.(2). Dependent and independent sources are treated the same way as in Thevenin's theorem.

Figure 2 |

Observe the close relationship between Norton's and Thevenin's theorems : R

_{N}= R_{Th}as in Equation.(1), and(3) |

This is essentially source transformation. For this reason, source transformation is often called Thevenin-Norton transformation.

Since V

_{Th}, I_{N}, and R_{Th}are related according to Equation.(3), to determine the Thevenin or Norton equivalent circuit requires that we find :- The open-circuit voltage v
_{oc}across terminals a and b. - The short-circuit current i
_{sc}at terminals a and b. - The equivalent or input resistance R
_{in}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

(4a) |

(4b) |

(4) |

the open-circuit and short-circuit test are sufficient to find any Thevenin or Norton equivalent, of a circuit which contains at least one independent source.

## Norton's Theorem Examples

For better understanding let us review the examples below :

**1. Find the Norton equivalent circuit of the circuit in Figure.(3) at terminals a-b.**

Figure 3 |

__Solution :__
We find R

_{N}in the same way we find R_{Th}in the Thevenin equivalent circuit. Set the independent sources equal to zero. This leads to the circuit in Figure.(4a), from which we find R_{N}. Thus,
To find I

_{N}, we short-circuit terminals a and b, as shown in Figure.(4b). We ignore the 5 Ω resistor because it has been short circuited. Applying mesh analysis, we obtain
From these equations, we obtain

Figure 4 |

Alternatively, we may determine I

_{N}from V_{Th/}R_{Th}. We obtain V_{Th}as the open-circuit voltage across terminals*a*and*b*in Figure.(4c). Using mesh analysis, we obtain
and

Hence,

as obtained previously. This also serves to confirm Equation.(4c) that R

_{Th}=*v*/_{oc}*i*= 4/1 = 4 Ω. Thus, the Norton equivalent circuit is as shown in Figure.(5)._{sc}Figure 5 |

**2.Using Norton's theorem, find R**

_{N}and of the cicuit in Figure.(6) at terminals*a-b*.Figure 6 |

__Solution :__
To find R

_{N}, we set the independent voltage source equal to zero and connect a voltage source of*v*= 1 V (or any unspecified voltage_{o}*v*) to the terminals. We obtain the circuit in Figure.(7a). We ignore the 4 Ω resistor because it is short-circuited. Also due to the short circuit, the 5 Ω resistor, the voltage source, and the dependent current source are all in parallel. Hence i_{o}_{x}= 0. At node a, i_{o}= 1v/5Ω = 0.2 A, and
To find I

_{N}, we short-circuited terminals*a*and*b*and find the current*i*, as indicated in Figure.(7b). Note from this figure that the 4 Ω resistor, the 10 V voltage source, the 5 Ω resistor, and the dependent current source are all in parallel. Hence,_{sc}
At node a, KCL gives

Thus

Figure 7 |

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