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:
Superposition Principle:
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
Example 2: Find voltage and current .
First, we solve this problem using node-voltage method. 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
Next, using superposition principle:
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 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 modeld as either a voltage source or a current source , and then find the and at its output port.
Thevenin's theorem: 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
Proof:
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.
The Norton's theorem can also be obtained by converting the voltage source model of the one-port network due to Thevenin's theorem in terms of in series with , to an equivalent current source model, in terms of in parallel with .
Example:
Also note that the Thevenin's voltage source and the Norton's current source can be converted into each other:
The voltage across a set of parallel branches of voltage source
in series with () is
Proof: (Homework)
This theorem can be augmented to include branches containing but no voltage source (), and branches contaning 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
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 can also be solved for , and of
the corresponding to get:
The top circuit in the following figure can be converted into either of the two equivalent circuits below.
The formed by , , and can
be converted to a Y, which can then be combined with to get
a Y (bottom left) with:
Example 0: (Homework)
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 1: In the circuit below, , , , . Find the value of current when is , , and . Moreover, find the value for for the desired current .
Method 1, KVL method First assume .
Applying KVL to the three loops: through , , and ,
through , , and , through , , and
:
Method 2, KCL method First assume .
Applying KCL to nodes a and b we get
Method 3, conversion
Find when . First convert the composed of
, and into a composed of , and :
Method 4, Thevenin's theorem
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 2: The circuit below, often used in some control system,
is composed of two voltages, two potentiometers, and a load resistor.
Assume:
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.
Method 1, Superposition theorem
Find caused by voltage , and then caused by voltage , then get .
Method 2, Thevenin's theorem
Remove , find open-circuit voltage and equivalent resistance , then find .