Norton's Theorem:
In some ways Norton's Theorem can be thought of as the opposite to "Thevenins Theorem", in that Thevenin reduces his circuit down to a single resistance in series with a single voltage. Norton on the other hand reduces his circuit down to a single resistance in parallel with a constant current source. Norton's Theorem states that "Any linear circuit containing several energy sources and resistances can be replaced by a single Constant Current generator in parallel with a Single Resistor". As far as the load resistance, RL is concerned this single resistance, RS is the value of the resistance looking back into the network with all the current sources open circuited and IS is the short circuit current at the output terminals as shown below.
Norton's equivalent circuit.
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The value of this "Constant Current" is one which would flow if the two output terminals where shorted together while the source resistance would be measured looking back into the terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section.
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To find the Nortons equivalent of the above circuit we firstly have to remove the centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.
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When the terminals A and B are shorted together the two resistors are connected in parallel across their two respective voltage sources and the currents flowing through each resistor as well as the total short circuit current can now be calculated as:
with A-B Shorted Out
If we short-out the two voltage sources and open circuit terminals A and B, the two resistors are now effectively connected together in parallel. The value of the internal resistor Rs is found by calculating the total resistance at the terminals A and B giving us the following circuit.
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Find the Equivalent Resistance (Rs)
Having found both the short circuit current, Is and equivalent internal resistance, Rs this then gives us the followingNortons equivalent circuit.
Nortons equivalent circuit.
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Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected across terminals A and B as shown below.
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Again, the two resistors are connected in parallel across the terminals A and B which gives us a total resistance of:
The voltage across the terminals A and B with the load resistor connected is given as:
Then the current flowing in the 40Ω load resistor can be found as:
which again, is the same value of 0.286 amps, we found using Kirchoff's circuit law in the previous tutorials.
Nortons Analysis Summary.
The basic procedure for solving Nortons Analysis equations is as follows:
- 1. Remove the load resistor RL or component concerned.
- 2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
- 3. Find IS by placing a shorting link on the output terminals A and B.
- 4. Find the current flowing through the load resistor RL
Thevenins Theorem : states that "Any linear circuit containing several voltages and resistances can be replaced by just a Single Voltage in series with a Single Resistor". In other words, it is possible to simplify any "Linear" circuit, no matter how complex, to an equivalent circuit with just a single voltage source in series with a resistance connected to a load as shown below. Thevenins Theorem is especially useful in analyzing power or battery systems and other interconnected circuits where it will have an effect on the adjoining part of the circuit.
Thevenins equivalent circuit.
As far as the load resistor RL is concerned, any "One-port" network consisting of resistive circuit elements and energy sources can be replaced by one single equivalent resistance Rs and voltage Vs, where Rs is the source resistance value looking back into the circuit and Vs the open circuit voltage at the terminals.
For example, consider the circuit from the previous section.
Firstly, we have to remove the centre 40Ω resistor and short out (not physically as this would be dangerous) all the emf´s connected to the circuit, or open circuit any current sources. The value of resistor Rs is found by calculating the total resistance at the terminals A and B with all the emf´s removed, and the value of the voltage required Vs is the total voltage across terminals A and B with an open circuit and no load resistor Rs connected. Then, we get the following circuit.
Find the Equivalent Resistance (Rs)
Find the Equivalent Voltage (Vs)
We now need to reconnect the two voltages back into the circuit, and as VS = VAB the current flowing around the loop is calculated as:
so the voltage drop across the 20Ω resistor can be calculated as:
VAB = 20 - (20Ω x 0.33amps) = 13.33 volts.
Then the Thevenins Equivalent circuit is shown below with the 40Ω resistor connected.
and from this the current flowing in the circuit is given as:
which again, is the same value of 0.286 amps, we found using Kirchoff´s circuit law in the previous tutorial.
Thevenins theorem can be used as a circuit analysis method and is particularly useful if the load is to take a series of different values. It is not as powerful as Mesh or Nodal analysis in larger networks because the use of Mesh or Nodal analysis is usually necessary in any Thevenin exercise, so it might as well be used from the start. However, Thevenins equivalent circuits of Transistors, Voltage Sources such as batteries etc, are very useful in circuit design.
Summary.
The basic procedure for solving Thevenins Analysis equations is as follows:
- 1. Remove the load resistor RL or component concerned.
- 2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
- 3. Find VS by the usual circuit analysis methods.
- 4. Find the current flowing through the load resistor RL.
Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense.
The strategy used in the Superposition Theorem is to eliminate all but one source of power within a network at a time, using series/parallel analysis to determine voltage drops (and/or currents) within the modified network for each power source separately. Then, once voltage drops and/or currents have been determined for each power source working separately, the values are all “superimposed” on top of each other (added algebraically) to find the actual voltage drops/currents with all sources active.
Let’s look at our example circuit and apply Superposition Theorem to it:
Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect. . .
. . . and one for the circuit with only the 7 volt battery in effect
When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks).
Analyzing the circuit with only the 28 volt battery
Analyzing the circuit with only the 7 volt battery
Super Imposing Voltage
Super Imposing current
Final Circuit
REVIEW
The Superposition Theorem states that a circuit can be analyzed with only one source of power at a time, the corresponding component voltages and currents algebraically added to find out what they’ll do with all power sources in effect.
To negate all but one power source for analysis, replace any source of voltage (batteries) with a wire; replace any current source with an open (break).
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