This is a Thevenin subcircuit If we apply KVL to this circuit we get the following equation: V_{T} + IR_{T} + Vx = 0 Vx = V_{T}  IR_{T} 

This is a Norton subcircuit If we apply KCL to this circuit we get the following equation: I_{N} = Vx/R_{N} + I Vx/R_{N} = I_{N}  I Vx = I_{N}R_{N}  IR_{N} 

Recalling from an early lecture, we stated that if two cicuits
yield the same mathematical equation then the two circuits are electrically equivalent as well. How can we make these two equations the same? These two circuits (& equations) are the same if the following is true: R_{T} = R_{N} V_{T} = I_{N}R_{N} 
This is a portion of the circuit you saw in problem 31. We know from 31 that this subcircuit reduces to the subcircuit shown at the right. In 31 we used source transformations to find the equivalent circuit, however now we will look at the current and voltage values at the output terminals. If the two circuits are equivalent then the voltage and current measured at the terminals should be identical. 

Calculating the OpenCircuit Voltage Both circuits above should have the same open circuit voltage as measured from the two terminals on the right: We will first calculate V_{opencircuit} or V_{oc} for the more complex circuit: Using current division: I_{8} = 6A[4/(4+12)] = 6A(1/4) = 3/2 A Note that the 4 and 12 Ohm resistors are in series since there is no current flowing to the output terminal. V_{oc} = I_{8}*4 Ohms = 3/2 A * 4 Ohms = 6 Volts Both the complex and the reduced circuits should have the same opencircuit voltage as measured from the two output terminals on the right: 

Calculating the ShortCircuit Current Now let's find the current across the output terminals when they are shorted by a wire. We will call this current I_{shortcircuit} or I_{sc} because it is the current which flows from one output terminal to the other when the two terminals are shorted together. 'Shorted' means the two terminals are connected by a wire. I_{sc} = I_{8} because there is no current in the 4 Ohm resistor. Since the rightmost 4 Ohm resistor is shorted by a wire the voltage across the 4 Ohm resistor is zero. If V=IR and V_{sc}=0 then I in the 4 Ohm resistor has to be zero. The 4 Ohm resistor has no effect on the circuit because no current flows through it. You could remove the rightmost 4 Ohm resistor if you wanted to for this step. Therefore applying Current Division: I_{8} = I_{sc} = 6A[4/(4+8)] = 6A(1/3) = 2A Both the complex and the reduced circuits should have the same shortcircuit current as measured from the two output terminals on the right: 

Now we calculate the OpenCircuit Voltage for the reduced Thevenin circuit. I=0 because the circuit is open. Current has no where to flow. KVL in the loop gives: 6V + 3I + V_{oc} = 0 Plugging in I=0: V_{oc} = 6 Volts This is the same opencircuit voltage found with the more complex circuit. NOTE: V_{oc} = V_{T} 

Now we calculate the ShortCircuit Current for the reduced Thevenin circuit. Here the loop current is equal to I_{sc} and V_{sc} = 0. Applying KVL to this circuit: 6V + 3I_{sc} + 0 = 0 6V = 3I_{sc} I_{sc} = 6V/3Ohms = 2A or I_{sc} = V_{T}/R_{T} This is the same shortcircuit current found with the more complex circuit. Since I_{N} = V_{T}/R_{T}, then: I_{sc} = I_{N} 

Summary: To find the Thevenin equivalent circuit for a linear circuit open up the output terminals and calculate V_{oc}. V_{T} = V_{oc}
To find R_{T} you need I_{sc} because I_{sc} = I_{N}Recall that R_{T} = V_{T}/I_{N} = V_{oc}/I_{sc} To find I_{sc}, short out the output terminals with a wire and calculate the current that flows from one terminal to the other. R_{T} = V_{oc}/I_{sc}Getting the Norton Equivalent circuit is done the same way since I_{N} = I_{sc}and R_{N} = R_{T}. 
Finding R_{T} or R_{N} directly:
To find the Thevenin/Norton Equivalent resistance you must set all independent sources to zero. Recall F(I) = V_{T}  R_{T}I. By setting all sources to zero in a subcircuit, this sets V_{T} in this equation to zero. If the terminals are left open that sets I to zero as well. Thus only R_{T} is left.Practically how is this done: First kill all independent sources (this means set voltage sources to zero volts and set current sources to zero amps). Then open up the output terminals. Calculate the resistance from one terminal to the other. 


Finding R_{T} in the first example:
We need to find R_{T} as seem between Terminal1 and Terminal2. The first step is to kill all independent sources. This is like unplugging the circuit and letting voltage sources be wires and current sources be open circuits. This "dead" circuit is shown on the right. Next we need to find the resistance of this "dead" circuit. Note that the 2 Ohm is in parallel with some open ciruits. Note that 2 Ohms in parallel with infinity = 2 Ohms. So the three resistors are in series. R_{T} = (5 + 2 + 8)Ohms = 15 Ohms 


Finding R_{T}  second example:
In this example the terminals are on the righthand side. We still need to find R_{T} as seem between Terminal1 and Terminal2. Again the first step is to kill all independent sources. The dead circuit is shown on the right. Next we need to find the resistance of this "dead" circuit. Note that the 10 Ohms is in series with an 'open' (effectively infinite resistance). This makes the 10 Ohm branch infinite resistance, so we can ignore this branch. We have two four Ohms resistors in parallel with each other. 4//4 = 2 (Note: '//' means 'in parallel') R_{T} = (8 + 4//4)Ohms = (8 + 2)Ohms = 10 Ohms 


Finding R_{T} in the first example using a different method:
Another way to find R_{T} or R_{N} is by exciting the dead circuit with an external independent source. The first step is to again 'kill' all independent sources. The second step is to place an independent source across the terminals, any current or voltage source will do. The third step involves finding the current and voltage at the terminals. Then R_{T} is = V_{in}/I_{in} Finding I_{in}: First we do KVL  1V + 5I_{in} + 2I_{in} + 8I_{in} = 0 1V = 15I_{in} I_{in} = 1/15 Amps R_{T} = V_{in}/I_{in} = 1 Volt/(1/15 Amps) = 15 Ohms This is the same answer found in the first example. 

The Practical Voltage Source:
A real world voltage source (i.e. practical voltage source) is modeled by what you see enclosed in the box. Note that a practical voltage source has some small but measureable internal resistance, R_{int}. If the load resistor R_{L} approaches zero, the current from the source can not exceed V_{S}/R_{int}. Thus current through the load does not approach infinity. With a perfect voltage source the current through the load can theoretically approach infinty. 

The Practical Current Source:
A real world current source (i.e. practical current source) is modeled by what you see enclosed in the box. Note that a practical current source has some small but measureable internal resistance, R_{int}. If the load resistor R_{L} approaches infinity (an open circuit), the current from the source into the load will approach zero. Voltage across the practical current source can not exceed I_{S}*R_{int}. Thus voltage across the load does not approach infinity. With a perfect current source the voltage across the load can theoretically approach infinty. 

Note that the current source and the voltage source are equivalent circuits. The voltage source is the Thevenin Circuit and the current source is the Norton Circuit. So if we never see a current source in the lab, why do we care about the current source on paper? The current source is a mathematical model which is can be used to simplify circuit analysis. However it may never be built in the real world. 