Thevenin - Norton Conversion

Article : Andy Collinson

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Thevenin to Norton Conversion

Thevenin to Norton

Sometimes circuit analysis can by solved easier by converting a Thevenin Circuit to a Norton equivalent Circuit. When converting from Thevenin to Norton the value of the Thevenin Resistance will be the same as the value of the Norton resistance.

To convert the above Thevenin circuit to a Norton circuit, the terminals A and B are short-circuited. The Norton Current is then the Thevenin Voltage V_{TH} divided by the Thevenin resistance, R which will be the same value in the Norton equivalent
circuit. Thus:

Norton to Thevenin

An Example

As an example in the circuit below I will solve the circuit to find the load current I_{L} flowing through RL, the load resistor.
As the load is variable (drawn with diagonal arrow) I_{L} will be calculated for four values of RL :- 0, 10k, 50k and 100k. I will
solve using Thevenin's Theorem, converting the circuit to the left of terminals A and B, then finally perform a Thevenin to
Norton conversion.

Now the equivalent Thevenin voltage is calculated. The load across A and B is removed but voltage sources E1 and E2 need to be taken into account. As the Thevenin source will be applied across A and B, with no load, R3 is considered open circuit and can be ignored. E1, R1, R2 and E2 are now in series. The value of this series current is:

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Thevenin to Norton Conversion

Thevenin to Norton

Sometimes circuit analysis can by solved easier by converting a Thevenin Circuit to a Norton equivalent Circuit. When converting from Thevenin to Norton the value of the Thevenin Resistance will be the same as the value of the Norton resistance.

To convert the above Thevenin circuit to a Norton circuit, the terminals A and B are short-circuited. The Norton Current is then the Thevenin Voltage V

I_{N} = |
V_{TH} |

R |

To convert from Norton to Thevenin the Norton theorem requires that the load across terminals A and B are open-circuited.
The Thevenin Voltage V_{TH} is then the product of the Norton current generated and circuit resistance. Thus:

V_{TH} = I_{N} * R

An Example

As an example in the circuit below I will solve the circuit to find the load current I

The Thevenin resistance is calculated. The load is removed, any voltage source short-circuited. The circuit is transformed as shown left.
The value of R_{TH} is:

R_{TH} = 30 + |
15 * 60 | = 42k |

( 15 + 60 ) |

Now the equivalent Thevenin voltage is calculated. The load across A and B is removed but voltage sources E1 and E2 need to be taken into account. As the Thevenin source will be applied across A and B, with no load, R3 is considered open circuit and can be ignored. E1, R1, R2 and E2 are now in series. The value of this series current is:

I = | E2 - E1 | = | 70 - 35 | = 0.4667mA |

(R1 + R2) | (15k + 60k) |

As current is the same at all points in a series circuit, the value of the Thevenin voltage, V_{TH} can now be calculated. The value of
V_{TH} will be E2 minus the voltage drop across R2. The voltage drop across R2 is I * R2, I being the previously derived current:

V_{TH} = E2 - I * R2 = 70 - 0.4667mA * 60k = 28 Volts

Now that both VTH and RTH have been found the Thevenin Equivalent circuit can be drawn, as shown left. Note that the circuit to the right of terminals A and B has been reduced by just one voltage source and one resistor.

Converting from Thevenin to Norton is now very easy the value of the Norton current source is:

I_{N} = |
42V | = 1 mA |

42k |

To solve the original question, the value of I_{L} for load resistors of 0 ohms, 10k, 50k and 100k. Using
current division the answers are:

I_{L} (for RL of 0 ohms) = 1 * |
42k | = 1 mA |

( 42k + 0 ) |

I_{L} (for RL of 10 kohms) = 1 * |
42k | = 0.808 mA |

( 42k + 10k ) |

I_{L} (for RL of 50 kohms) = 1 * |
42k | = 0.457 mA |

( 42k + 50k ) |

I_{L} (for RL of 100 kohms) = 1 * |
42k | = 0.296 mA |

( 42k + 100k ) |