Thevenin Theorem This theorem is very conceptual. If we think deeply about an electrical circuit, we can visualize the statements made in Thevenin theorem. Suppose we have to calculate the current through any particular branch in a circuit. This branch is connected with rest of the circuits at its two terminal. Due to active sources in the circuit, there is one electric potential difference between the points where the said branch is connected. The current through the said branch is caused by this electric potential difference that appears across the terminals. So rest of the circuit can be considered as a single voltage source, that's voltage is nothing but the open circuit voltage between the terminals where the said branch is connected and the internal resistance of the source is nothing but the equivalent resistance of the circuit looking back into the terminals where, the branch is connected. So the Thevenin theorem can be stated as follows,

When a particular branch is removed from a circuit, the open circuit voltage appears across the terminals of the circuit, is Thevenin equivalent voltage and, The equivalent resistance of the circuit network looking back into the terminals, is Thevenin equivalent resistance. If we replace the rest of the circuit network by a single voltage source , then the voltage of the source would be Thevenin equivalent voltage and internal resistance of the voltage source would be Thevenin equivalent resistance which would be connected in series with the source as shown in the figure below.

ere two resistors R₁ and R₂ are connected in series and this series combination is connected across one voltage source of emf E with internal resistance R_i as shown. One resistive branch of R_L is connected across the resistance R₂ as shown. Now we have to calculate the current through R_L.

First, we have to remove the resistor R_L from the terminals A and B.

Second, we have to calculate the open circuit voltage or Thevenin equivalent voltage V_T across the terminals A and B.

The current through resistance R₂,

Hence voltage appears across the terminals A and B i.e.

Third, for applying Thevenin theorem, we have to determine the Thevenin equivalent electrical resistance of the circuit, and for that; first we have to replace the voltage source from the circuit, leaving behind only its internal resistance R_i. Now view the circuit inwards from the open terminals A and B. It is found the circuits now consist of two parallel paths - one consisting of resistance R₂ only and the other consisting of resistance R₁ and R_i in series.

Thus the Thevenin equivalent resistance R_T is viewed from the open terminals A and B is given as. As per Thevenin theorem, when resistance R_L is connected across terminals A and B, the network behaves as a source of voltage V_T and internal resistance R_T and this is called Thevenin equivalent circuit. The current through R_L is given as,

Norton’s Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single current source and parallel resistance connected to a load. Just as with Thevenin’s Theorem, the qualification of “linear” is identical to that found in the Superposition Theorem: all underlying equations must be linear (no exponents or roots).

Contrasting our original example circuit against the Norton equivalent: it looks something like this:

. . . after Norton conversion . . .

Remember that a current source is a component whose job is to provide a constant amount of current, outputting as much or as little voltage necessary to maintain that constant current.

As with Thevenin’s Theorem, everything in the original circuit except the load resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to Thevenin’s Theorem are the steps used in Norton’s Theorem to calculate the Norton source current (INorton) and Norton resistance (RNorton).

As before, the first step is to identify the load resistance and remove it from the original circuit:

Then, to find the Norton current (for the current source in the Norton equivalent circuit), place a direct wire (short) connection between the load points and determine the resultant current. Note that this step is exactly opposite the respective step in Thevenin’s Theorem, where we replaced the load resistor with a break (open circuit):

With zero voltage dropped between the load resistor connection points, the current through R1 is strictly a function of B1‘s voltage and R1‘s resistance: 7 amps (I=E/R). Likewise, the current through R3 is now strictly a function of B2‘s voltage and R3‘s resistance: 7 amps (I=E/R). The total current through the short between the load connection points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps becomes the Norton source current (INorton) in our equivalent circuit:

Remember, the arrow notation for a current source points in the direction opposite that of electron flow. Again, apologies for the confusion. For better or for worse, this is standard electronic symbol notation. Blame Mr. Franklin again!

To calculate the Norton resistance (RNorton), we do the exact same thing as we did for calculating Thevenin resistance (RThevenin): take the original circuit (with the load resistor still removed), remove the power sources (in the same style as we did with the Superposition Theorem: voltage sources replaced with wires and current sources replaced with breaks), and figure total resistance from one load connection point to the other:

Now our Norton equivalent circuit looks like this:

If we re-connect our original load resistance of 2 Ω, we can analyze the Norton circuit as a simple parallel arrangement:

As with the Thevenin equivalent circuit, the only useful information from this analysis is the voltage and current values for R2; the rest of the information is irrelevant to the original circuit. However, the same advantages seen with Thevenin’s Theorem apply to Norton’s as well: if we wish to analyze load resistor voltage and current over several different values of load resistance, we can use the Norton equivalent circuit again and again, applying nothing more complex than simple parallel circuit analysis to determine what’s happening with each trial load.

with Matildo,