# Superposition Theorem

#### steeve_wai

Joined Sep 13, 2007
47
why can this theorem be applied only to linear functions such as voltage and current,and not power which is non-linear.please explain with equations.thank you.

#### recca02

Joined Apr 2, 2007
1,212
its not ' non linear functions' that they r not applicable to IIRC (correct me if i m wrong).
its not applicable to non linear circuits.

#### Eduard Munteanu

Joined Sep 1, 2007
86
If you model the current entering/exiting a node as a function, then you can say that $$f(\sum_{k=1}^n{I_k}) = \sum_{k=1}^n{f(I_k)}$$ iff f is linear, where $$I_k$$ is the current contributed by source k. This is easy to prove if one writes $$f(x) = ax + b$$. It's also easy to extend it to voltages.

How to prove it the other way eludes me at the moment.

$$f(E) = {1 \over R} E^2$$, which is for power, certainly doesn't obey the above relationship.

#### BlackBox

Joined Apr 22, 2007
20
Sorry to appoint it Eduard, but you are wrong: $$f(x) = a x + b[\tex] is not a linear function if you want to speak of the superposition theorem. A linear function is a function between vector fields (a scalar field can be a vector field too). A function is linear if it follows the following, simple, rules: \( f(ax)= a \cdot() f (x)[\tex] \( f(x_1+x_2)=f(x_1)+f(x_2)[\tex] given x in a vector field Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals). The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation: \( f(x) = x^2 [\tex] \( f(x_1 + x_2) = (x_1 + x_2)^2 [\tex] \( f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex] \( f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] Q.E.D. It's not easy to understand if you are not confortable with vector fields, ask if you have some problems... Holla$$\)\)\)\)\)\)