# Why is a circuit of linear elements itself a linear circuit?

First on what I mean by linear element and linear circuit.

A linear element has a voltage that is proportional to it's current. A resistor's voltage is proportional to it's current and in the frequency domain capacitors and inductors are also linear elements. This element has a property of superposition such that the output of an input is equivalent to the sum of the outputs of two smaller inputs that sum to the original input.

A linear circuit has a different definition. It is defined by satisfying the superposition theorem, different than the property of superposition mentioned above for the linear element. The theorem states that the voltage or current anywhere on the circuit due to multiple input sources is the same as the sum of the responses which would occur if only one source was turned on at a time. This is very different from the meaning of a linear element whose superposition property only applied to inputs at one location instead of multiple inputs in different locations.

Which brings to my question: Why does a circuit of linear elements create a linear circuit?

Note: The current answers say that linear elements imply a linear circuit without explaining why. Or at least not clearly enough for me to understand.

• Resistors and sources are defined by linear, time independent relationships. Apr 20 '16 at 19:36
• @CuriousOne Can you clarify? Apr 20 '16 at 19:47
• Look at the definitions. Apr 20 '16 at 20:22
• so a resistor's Voltage is proportional to it's current. V=iR. And a source maintains a constant voltage or current. So these are linear, time independent relations. If I put combinations of them in a circuit, they are time independent as well. But why must they be linear? Apr 22 '16 at 5:46
• @CuriousOne Forgot to alert you. Apr 22 '16 at 5:47

Well, you could use the mathematical definition of linear.

Definition (Linear) - An object $$H$$ is called linear if it satisfies the following principle of superposition:

$$\alpha H(x) + \beta H(y) = H(\alpha x + \beta y)$$

If $$H$$ is described as a linear operation (in this case the operation of resistances in a circuit) satisfy superposition.

Now, if you take passive elements such as an inductor or capacitor, you'll see that they are not proportionally/linearly related to the current:

$$V_r = IR, \hspace{6pt} V_i = L\frac{dI}{dt}, \hspace{6 pt} V_c = \frac{1}{c}\int I\hspace{2pt} \mathrm{d}x$$

However, these also satisfy the principle of superposition, because the derivative and integral operators are linear operators themselves. So a circuit can be analysed in terms of the properties of its elements (linear or non-linear) and solved accordingly.

• But with that definition a voltage source isn't linear? Vs(i)=Vo+0*i. Vs(1)+Vs(2)=2*Vo does not equal Vs(1+2)=Vo. Also, although we know a resistor and capacitor's own currents and voltages are linear, how do we know the resistor's voltage is also linear with respect to the capacitor's current in any arbitrary circuit? May 4 '16 at 1:24

Imagine an arbitrarily complex network made up solely of linear elements.

Now take four terminals anywhere on the network, which will act as our two ports. The network itself is the famous black box of 2 port network.

Now apply Kirchhoff's laws to all possible loops in the network. You will get lot of linear equations, solving which will give you the value of Voltage and Current at every node in the network.

Since the four terminals themselves are connected to some nodes their values should also be included into the set of linear equations (as variables $$V_1 , I_1, V_2, I_2$$).

So after solving you will get a linear relationship between $$V_1 , I_1, V_2, I_2$$, i.e, the port current and voltage values.

This is the definition of a linear network. Hence proved.

If you have all linear relationships, the outcome must be a linear one. If you have apples, do anything to them, you will still have apples

• but how do you know there isn't some elaborate circuit of linear elements where two properties in the circuit aren't linear to each other? Apr 30 '16 at 18:18
• That again depends on the circuit roobee, if you have components which are time dependent in their behaviour, for example a capacitor, then your overall circuit behaviour will not be linear, Apr 30 '16 at 18:26
• That's not exactly true. If you cut apples down to each atom, you don't really have an apple anymore. Apr 30 '16 at 19:07