Thevenin and Norton Networks

Article : Andy Collinson

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_{TH }and R_{TH}

To find the equivalent Thevenin resistance of the circuit. R_{TH} the load is first removed and any circuit voltages are short circuited.
The resistance is then calculated. The Thevenin voltage is found by first removing the circuit load and then working out the voltage across point
A and B in the circuit.

An Example

A demonstration of the thevenin technique to find current I_{1} in the diagram below:

Norton's Theorem

The Norton theorem converts an ordinary circuit to an equivalent parallel circuit which is a current source in parallel with a resistor. The technique is similar to the thevenin theorem and two points in a circuit must be defined, this is where the analysis will take place.

_{TH }and R_{TH}

The value of Vth is found by either measuring (if you don't know what's in the circuit) or be using circuit analysis. To find Rth ( with load removed) short circuit voltage supplies, open current sources and calculate the equivalent resistance.

An Example

Norton's theorem is demonstrated to find the current flowing through the 50 ohm resistor, I_{1} in the diagram below :

The current through the load, I, can now be found using the current division rule :

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Two powerful circuit analysis techniques are Thevenin's theorem and Norton's theorem. Both theories convert a complex circuit to a simpler series or parallel equivalent circuit for easier analysis. Analysis involves removing part of the circuit across two terminals to aid calculation, later combining the circuit with the Thevenin or Norton equivalent circuit.

Thevenin's TheoremThe top left diagram represents the circuit for analysis at terminals A and B. The top right hand circuit is the Thevenin equivalent circuit, a voltage source Vth with a series resistance, Rth. The bottom left diagram is the same circuit driving a load. The load is NOT included in the thevenin equivalent circuit and must be separated, this is why the terminals are marked A and B.

Value of VTo find the equivalent Thevenin resistance of the circuit. R

An Example

A demonstration of the thevenin technique to find current I

First, the circuit to the right of points A and B is converted to a Thevenin source and resistance. The components to the left, the 30v battery and 10 ohm resistor are removed. The Thevenin voltage is then the equivalent voltage across terminals A and B becomes:

V_{TH} = |
40 * 10 | = 13.3333 |

30 |

The circuit is now converted to find the Thevenin Resistance (seen looking into the circuit to the right of points A and B). All voltage sources are short circuited and the equivalent Thevenin Resistance will be the value of the 10 ohm resistor in parallel with the 20 ohm resistor.

R_{TH} = 10||20= |
20 * 10 | = 6.6667 |

(20+10) |

Knowing both the equivalent Thevenin Voltage and Thevenin Resistance, the value of current I_{1} can now be solved. I1 flows in
series and from ohm's law will be the voltage difference ( 30 V source - VTH) divided by total resistance:

I_{1} = |
30 - V_{TH } |
= | 30 - 13.3333 | = 1 Amp |

(10 + R_{TH}) |
(10 + 6.6667) |

The example above is messy numerically, and it is important to use sufficient decimal places to be accurate. Instead of decimals, improper fractions can be used, which are sometimes easier to calculate.

Norton's Theorem

The Norton theorem converts an ordinary circuit to an equivalent parallel circuit which is a current source in parallel with a resistor. The technique is similar to the thevenin theorem and two points in a circuit must be defined, this is where the analysis will take place.

As with Thevenin, the equivalent circuit is a current generator In and norton equivalent resistance, Rn. These must be worked out to use the Norton theorem. The analysis points using Norton are short circuited, whereas using the Thevenin Method they are open circuit.

Value of VThe value of Vth is found by either measuring (if you don't know what's in the circuit) or be using circuit analysis. To find Rth ( with load removed) short circuit voltage supplies, open current sources and calculate the equivalent resistance.

An Example

Norton's theorem is demonstrated to find the current flowing through the 50 ohm resistor, I

The points A and B is where the Norton conversion takes place, the rightmost 50 ohm (load) resistor is first removed, and diagram redrawn with terminals A and B are short circuited, see below:

First the Norton Resistance R_{N} is calculated. Any current sources are short circuited and any voltage sources are drawn open
circuit as shown in the left diagram. The resistance in the circuit is now a 50 ohm resistor in series with a 100 ohm resistor and another
50 ohm in parallel. The Norton Resistance is:

R_{N} = 50 + |
50 * 100 | = 83.3333 Ohm |

( 50 + 100 ) |

First the total current I_{t} is calculated using ohm's law, this is the battery voltage divided by total resistance. The resistance in
the circuit is now a 50 ohm resistor in series with a 100 ohm resistor and another 50 ohm in parallel. (100 ohm in parallel with 50 ohm is
33.3333.) The Norton Resistance R_{N} is:

I_{t} = |
100 | = 1.2 Amp |

( 50 + 33.3333) |

Now that I_{t} is known the circuit can be redrawn as shown left to find the value of the equivalent Norton Current.
Using the current division rule, I_{N} is worked out:

I_{N} = |
1.2 * 100 | = 0.8 Amp |

( 50 + 100) |

The Norton equivalent circuit can now be completed with the right hand 50 ohm resistor included:

The current through the load, I, can now be found using the current division rule :

I = | 0.8 * 83.333 | = 0.5 Amp |

( 50 + 83.3333) |

The results can be verified using the Thevenin method or a circuit simulation.