Thevenin's Theorem and Norton's Theorem

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• Thevenin’s Theorem is a way to reduce a network to an equivalent circuit composed of a single voltage source, series resistance, and series load.
• Steps to follow for Thevenin’s Theorem:
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• (1) Find the Thevenin source voltage by removing the load resistor from the original circuit and calculating voltage across the open connection points where the load resistor used to be.
• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Thevenin equivalent circuit, with the Thevenin voltage source in series with the Thevenin resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for series circuits.

• Norton’s Theorem is a way to reduce a network to an equivalent circuit composed of a single current source, parallel resistance, and parallel load.
• Steps to follow for Norton’s Theorem:
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• (1) Find the Norton source current by removing the load resistor from the original circuit and calculating current through a short (wire) jumping across the open connection points where the load resistor used to be.
• (2) Find the Norton resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points.
• (3) Draw the Norton equivalent circuit, with the Norton current source in parallel with the Norton resistance. The load resistor re-attaches between the two open points of the equivalent circuit.
• (4) Analyze voltage and current for the load resistor following the rules for parallel circuits.
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 Since Thevenin’s and Norton’s Theorems are two equally valid methods of reducing a complex network down to something simpler to analyze, there must be some way to convert a Thevenin equivalent circuit to a Norton equivalent circuit, and vice versa (just what you were dying to know, right?). Well, the procedure is very simple.

You may have noticed that the procedure for calculating Thevenin resistance is identical to the procedure for calculating Norton resistance: remove all power sources and determine resistance between the open load connection points. As such, Thevenin and Norton resistances for the same original network must be equal. Using the example circuits from the last two sections, we can see that the two resistances are indeed equal:

Considering the fact that both Thevenin and Norton equivalent circuits are intended to behave the same as the original network in supplying voltage and current to the load resistor (as seen from the perspective of the load connection points), these two equivalent circuits, having been derived from the same original network should behave identically.

This means that both Thevenin and Norton equivalent circuits should produce the same voltage across the load terminals with no load resistor attached. With the Thevenin equivalent, the open-circuited voltage would be equal to the Thevenin source voltage (no circuit current present to drop voltage across the series resistor), which is 11.2 volts in this case. With the Norton equivalent circuit, all 14 amps from the Norton current source would have to flow through the 0.8 Ω Norton resistance, producing the exact same voltage, 11.2 volts (E=IR). Thus, we can say that the Thevenin voltage is equal to the Norton current times the Norton resistance:

So, if we wanted to convert a Norton equivalent circuit to a Thevenin equivalent circuit, we could use the same resistance and calculate the Thevenin voltage with Ohm’s Law.

Conversely, both Thevenin and Norton equivalent circuits should generate the same amount of current through a short circuit across the load terminals. With the Norton equivalent, the short-circuit current would be exactly equal to the Norton source current, which is 14 amps in this case. With the Thevenin equivalent, all 11.2 volts would be applied across the 0.8 Ω Thevenin resistance, producing the exact same current through the short, 14 amps (I=E/R). Thus, we can say that the Norton current is equal to the Thevenin voltage divided by the Thevenin resistance:

This equivalence between Thevenin and Norton circuits can be a useful tool in itself, as we shall see in the next section.

• REVIEW:
• Thevenin and Norton resistances are equal.
• Thevenin voltage is equal to Norton current times Norton resistance.
• Norton current is equal to Thevenin voltage divided by Thevenin resistance.

Source Transformation

Source transformation is simplifying a circuit solution, especially with mixed sources, by transforming a voltage into a current source, and vice versa. Finding a solution to a circuit can be difficult without using methods such as this to make the circuit appear simpler.

The source transformation of a circuit is the transformation of a power source from a voltage source to a current source, or a current source to a voltage source.

In other words, we transform the power source from either voltage to current, or current to voltage. 

The direction of the arrow of the current source follows the positive sign of the voltage source.

EXAMPLE

In this circuit, we should identify the voltage at the 4Ω resistor. We will start transforming at the right side of the circuit using simple ohms law.

V= 2A (2Ω)

V= 4 V

The resulting circuit would be:

Combine the series impedance’s.

Then, transform it again.

I=V/R

I= 4/ 2-j1

I= 1.788∟26.56 A

Combine the parallel impedances.

Z= j3(2-j1)/ j3+(2-j1)

Z= 2.37 ∟18.43 Ω

Then transform again so that we can subtract the voltage source and finally obtain V.

V= (1.788∟26.56) (2.25+j0.75)

V= 2.998 + j2.998

V= V1-V2

V= (2.998 +j2.998)-(10)

V= 7.6168 ∟156.82 V

We can now use voltage division to obtain V.

V=(7.616∟156.82)(4)/4+ (2.25+j0.75)

V= 4.84 ∟149.97 V

Superposition Theorem

The Superposition Theorem states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or current through) that element due to each independent source acting alone.

The principle of Superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. However to apply the superposition principle, we must keep two things in mind:

1. We consider one independent source at a time while all other independent sources are turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and every current sources by 0 A (or an open circuit). This way we obtain a simpler and more manageable circuit.

2. Dependent sources are left intact because they are controlled by circuit variables.

Steps to Apply Superposition:

1. Turn off all independent sources except one source. Find the output voltage r current due to that active source using the techniques.

2. Repeat step 1 for each of the other independent sources.

3. Find the total contribution by adding algebraically all the contributions due to independent sources.

Mesh Analysis

Another method of analyzing circuits and the counterpart of nodal analysis is the mesh analysis. Also known as loop analysis or the mesh-current method. A mesh is a loop which does not contain any other loops within it. The current through a mesh is known as mesh current.

The steps in determining mesh currents:

• Assign the meshes found in the circuit as mesh currents i1, i2, … , in.

• Apply KVL to each of the assigned meshes to form equations. Use Ohm’s law (V=IR) to express the voltages in terms of the mesh currents.

• Solve the formed equations from step 2 to get the value of the unknown mesh currents. 

Example circuit:

Nodal Analysis

Nodal analysis is one of the many methods in solving and finding a specific value of a parameter in electronic circuit analysis. The aim of using nodal analysis is to determine the voltage in each node that’s relative to the reference node, which is the ground GND where voltage is equal to 0. This means that all the other nodes present in the circuit are referred to as the non-reference nodes; the ones that has voltage you are trying to solve for. Depends, of course, if you do need the voltage present in them or not.

Steps on how to determine node voltages:

 

• Determine the nodes of the circuit, and then select a node as the reference node (ground GND). Then assign the non-reference nodes to voltages V1, V2, Vx, or whatever you feel comfortable with.

• Apply KCL (Kirchhoff’s Current Law) to each of the non-reference nodes. Use Ohm’s law (V=IR) to express the branch currents in terms of node voltages. I use the shortcut method, though the same principles are still applied.

• Solve the resulting simultaneous equations formed from the non-reference nodes to obtain the unknown node voltages.

• Always remember that the number of equations formed should be equal to the number of unknowns. 

Example circuit:

 

Impedance and Admittance

The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms.

In some ways, you could say that the impedance acts the same as the resistance in a circuit, but not completely.

The admittance Y is the reciprocal of impedance, measured in siemens.

 

Resistors, capacitors, and inductors are the three elements that cause impedance. Though you have to convert the capacitance C and the inductance L to get their equivalent impedances Z, by applying these formulas: