A generic system with input (stimulus, cause) and output (response, effect) can be described mathematically as a function . The function is linear if it satisfies the following:
When there exist multiple energy sources in the circuit, any voltage and current in the circuit can be found as the algebraic sum of the corresponding values obtained by assuming only one source at a time, with all other sources turned off:
As superposition principle only applies to linear functions, it cannot be applied to nonlinear functions such as power (e.g., or ).
However, note that superposition principle does not apply to any variable
nonlinearly related to the energy sources, such as power:
Example 1: The previous example can also be solved by superposition theorem.
First, we turn the voltage source of 20V off (short-circuit with 0V), and get
Thevenin's theorem and Norton's theorem
In principle, all currents and voltages of an arbitrary network of linear components and voltage/current sources can be found by either the loop current method or the node voltage method.
However, if only the current and voltage associated with one particular component such as a resistor are of interest, it is unnecessary to find voltages and currents elsewhere in the circuit. Instead, we can ``pull'' the component out and treat it as the load of the rest of the circuit, which can be modeled as either a Thevenin voltage source or a Norton current source , and then find the and at its output port.
Any one-port (two-terminal) network of resistance elements and energy sources is equivalent to an ideal voltage source in series with a resistor , where
We see that as far as the port voltage and current associated with the load are concerned, the one-port network is equivalent to an ideal voltage source , the open-circuit voltage across the port, in series with an internal resistance , which can be obtained as the ratio of the open-circuit voltage and the short-circuit current.
Any one-port (two-terminal) network of resistance elements and energy sources is equivalent to an ideal current source in parallel with a resistor , where
Proof: The proof of this theorem is in parallel with the proof of Thevenin's theorem.
We see that as far as the port voltage and current associated with the load are concerned, the one-port network is equivalent to an ideal current source , the short-circuit current through the port, in parallel with an internal resistance , which can be obtained as the ratio of the open-circuit voltage and the short-circuit current.
In fact, Thevenin's theorem and Norton's theorem are equivalent, as
one can be converted into the other. The internal resistances in both
theorems are the same , and the voltage source in
series with in Thevenin's theorem can be converted to a current
in parallel with in
Norton's theorem. In either case, we can find the internal resistance
The voltage across a set of parallel branches of voltage source
in series with () is
This theorem can be augmented to include branches containing but no voltage source (), and branches containing a current source . If a resistance is in series with , it can be neglected (as it does not affect the current). If a resistance is in parallel with , they can be converted into a voltage source in series with .
The configuration can be converted to and vice versa. We
relate the and by realizing that the resistance
between terminals a and b of should be equal to that between the
same two terminals of :
Given , and of a , the three equations
above can be solved for , and of the corresponding .
For example, subtracting the 3rd equation from the sum of the first two,
we get expression for . The solutions are:
Reversely, given , and of a , the same three
equations above can also be solved for , and
of the corresponding to get:
The top circuit (a bridged T-network) in the following figure can be converted into either of the two equivalent circuits below.
Example 0: Model the circuit in part (a) by Thevenin's theorem (b) and Norton's theorem (c).
Example 1: Find voltage and current .
Any one of the three methods can be used to solve the circuit:
Assume the currents (left branch), and (right branch)
all leave the top node, where the voltage is (with respect to the
bottom treated as ground). By KCL, we have
Example 2: (Homework)
In general, the conversion from to is more useful as is easier to analyze than . For example, the circuit in (a) below can be converted to that in (b) to find all the currents in the circuit:
Assume , , , , , , , find the currents , , , , , and .
Example 3: In the circuit below, , , , . Find the value of current when . Then resolve the problem for and . Moreover, find the value for for the desired current .
Consider three independent loops with clockwise loop currents:
Assume the bottom node is ground with 0 volt, then the voltage at
the top node is known to be . Now we apply KCL to nodes
a and b to get two equations:
To find for the current through it to be , we
replace the current through in the two
equations above by the desired :
To find through , we first convert the
composed of , and into a composed of ,
For any of the methods above, the problem needs to be resolved for and , and it is hard to find a value of given the require current .
Solve the problem using Thevenin's theorem by the following steps:
Specifically, we remove as the load of a network composed of all other
resistors , , , and the voltage source , then
apply Thevenin's theorem to find the open-circuit voltage between the two
terminals a and b:
Example 4: The circuit below, often used in some control system,
is composed of two voltages, two potentiometers, and a load resistor.
We denote the current through by , and the voltage at the left and right ends of by and , respectively, with respect to the bottom wire treated as the ground.
Find caused by voltage , and then caused by voltage , then get .
Remove , find open-circuit voltage and equivalent resistance , then find .