TLO 11 State and apply the following basic theorems for simplifying the analysis of linear resistive networks: superposition,
Thevenin's, Norton's, maximum power transfer and reciprocity.
ELO 11.1 State the superposition theorem, and emphasize that it applies only to linear current-voltage relations.
ELO 11.2 Illustrate, by working examples, the application of the theorem in solving electric circuits containing
multiple sources of the same type and of both types.
ELO 11.3 State Thevenin's Theorem, and find the Thevenin equivalent of at least three electric circuits.
ELO 11.4 State Norton's theorem, and find the Norton equivalent of at least three electric circuits.
ELO 11.5 Convert between the Thevenin and Norton equivalent parameters.
ELO 11.6 State the maximum power transfer theorem and illustrate its use by at least two examples.
ELO 11.7 State the reciprocity theorem and illustrate its use by application to one example.
SUPERPOSITION THEOREM
- Theorem: The current through, or voltage across, an element in a linear bilateral network is equal to the algebraic sum of the
currents or voltages produced independently by each source.
- A source acting independently means that each source acts alone, all other sources being made zero
- For each source acting alone, find the current in each element.
- Keep track of each element current associated with each source by using prime or double prime or triple prime symbols.
- I Any element = I (i.e. due to Source 1 alone) + I (i.e. due to Source
2 alone) + I (i.e. due to Source 3 alone)
- Sources being made zero means:
- Advantage of using the Superposition Theorem
- With one source acting at a time, we always end up with the remainder of the circuit being series-parallel resistors
- Series-parallel is the first solution method that we learned, and therefore the most familiar.
Disadvantages of using the Superposition Theorem
- If we have N sources, we have to solve N individual circuits (i.e. with each of the sources acting alone)
- There is a chance for making an algebraic mistake if we do not keep the same current reference direction each time.
EXAMPLE OF USING SUPERPOSITION THEOREM
(similar, but not identical to, Example 9.3)
Find the current I2 using the Superposition Theorem. (with the reference
direction for I2 being down in the resistor)
- The 36 volt Voltage Source acting alone:
The only active part of the circuit remaining is a single loop. It is easy to
solve
- The 9 amp Current Source acting alone:
By the current divider rule:
I Any element = I (due to Source 1 alone) + I (due to Source 2 alone)
I 2 = I2 (due to 36 volt source alone) + I2 ( due to 9 A source alone)
I2 = - 2 A + 6 A = 4 A
P = I2 R watts
- The total current squared is not equal to the sum of the square of each individual current.
- Therefore we cannot use the Superposition Principle to calculate the power dissipated in a resistor, because the relationship
between current and power is not linear.
- We must always find the total current in a resistor before attempting to calculate the power dissipated in the resistor.
THEVENIN'S THEOREM
- Theorem: Any two - terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a voltage source
and series resistor, as shown in the figure below.
Thevenin equivalent circuit
- How could Thevenin have stated such a Theorem, true for all linear bilateral networks in existence over hundreds of years ???
He saw one of two things.
He wrote standard mesh equations for the original network.
He understood the method of solving the mesh equations, which is:
- elimination of variables
- back substitution
- He knew that when he got down to the last variable, the equation would have the form
y = - m x + b
Terminal voltage = - RTH I + ETH
VT = ETH - RTH I
Or
- He saw that it was possible to do a series of source conversions, together with series-parallel combinations of resistances, to end
up with only one voltage source ETH and one resistor RTH .
- This is the most straightforward method of finding the Thevenin Equivalent Circuit.
- We will illustrate this straightforward method using the last circuit from the Superposition method
Linear bilateral
network (it could be a network of any structure)
Convert the 36 volt - 12 ohm series path to current source form
Combine the resistors which are in parallel,
and combine the current sources which are in parallel
Convert the 6 amp
current source which is in parallel with the resistor of 4 ohms to a voltage source form.
This automatically gives the ETH and the RTH.
Classical Method of Finding RTH and ETH
- Preliminary:
- Sometimes we may not want to reduce all of the network to a Thevenin Equivalent, so we should first
- Isolate, or set apart, the portion of the network for which the Thevenin equivalent is to be found
This isolation has to be done at two points (only)
- Identify, or mark the terminals of the remaining two - terminal network.
RTH
The logic for the steps is shown on the diagram
- Set all sources to zero and then find the resultant resistance between the two marked terminals (voltage sources are replaced by
short circuits and current sources by open circuits)
ETH
- The logic for the steps is shown on the diagram
- Return all sources to their original position
- KVL in the right circuit says:
ETH - IRTH = Vab ,and with I = 0 amps (open circuit)
ETH = Vab on open circuit (called VOC)
NORTON 'S THEOREM
Any two - terminal linear bilateral dc network can be replaced by an equivalent circuit consisting of a current source and a resistor
in parallel as shown in the figure below:
Norton Equivalent Circuit
- How could Norton have stated such a Theorem, true for all linear bilateral networks over hundreds of years ??? He saw one
of two things.
- He believed Thevenin's Theorem, and simply knew how to do a source conversion.
Or
- He saw that it was possible to do a series of source conversions, together with series-parallel combinations of resistances, to
end up with only one current source IN and one resistor RN.
- This is the most straightforward method of finding the Norton Equivalent Circuit.
- We will illustrate this straightforward method by using the first three steps of our Thevenin process.
Linear bilateral network (it could be a network of any structure)
Convert the 36 volt - 12 ohm series path to current source form.
Combine the resistors which are in parallel,
and combine the current sources which are in parallel
This automatically gives the IN and the RN.
Classical Method of Finding RN and IN
- Preliminary:
- Sometimes we may not want to reduce all of the network to a Thevenin Equivalent, so we should first
- Isolate, or set apart, the portion of the network for which the Norton equivalent is to be found
This isolation has to be done at two points (only)
- Identify, or mark the terminals of the remaining two - terminal network
- The logic for the steps is
shown on the diagram
- Set all sources to zero, and find the resultant resistance between the two marked terminals (voltage sources are replaced by
short circuits and current sources by open circuits)
IN
- Return all sources to their original position.
Short circuit the terminals
- Current divider in right circuit says:
MAXIMUM POWER TRANSFER THEOREM
A source - practical voltage source
A source - large networkas a Thevenin's equivalent
- If we can choose the load, what value of load resistor RL will cause the source to deliver the maximum power of which it is
capable
- Textbook uses trial values of RL to show that maximum power comes to the load when RL = R (the source resistor)
- Circuit analysis + calculus will not hurt
and PL = I2 RL so
- From the calculus, to find the maximum value of PL
(You will be taking the derivative of a quotient)
CONVENIENCE EXPRESSIONS
- Knowing that RL = R, everything can be found from basic circuit analysis
- We can also use fundamentals to develop convenience formulas
PL MAX = I2 RL, and substituting
EFFICIENCY
When the load power is PL MAX, RL = R
MAXIMUM POWER TO LOAD - WHO WANTS IT
- E and R represent the microphone, amplifiers, and signal transmission line to the antenna
- RL represents the antenna as a load (RL = R)
- Want maximum power to the antenna, so they will get maximum radio waves and maximum listener coverage
- Will accept the 50% efficiency
My house: NO
- E and R represent the NB Power generators and transmission lines
- RL represents my house load (RL R)
- If I got maximum power from Point Lepreau (half of 680 MW), it would melt my house down
- Could not afford the power bill at 50% efficiency.