# Linearity Property

A linear circuit is one whose output is linearly related (or directly proportional) to its input.
Linearity is the property of an element describing a linear relationship between cause and effect. Although the property applies to many circuit elements, we shall limit its applicability to resistors in this chapter. The property is a combination of both the homogeneity (scaling) property and the additivity property.
The homogeneity property requires that if the input (also called the excitation) is multiplied by a constant, then the output (also called the response) is multiplied by the same constant. For a resistor, for example, Ohm's law relates the input i to the output v, $$v = iR \tag{1}$$ If the current is increased by a constant k, then the voltage increases correspondingly by k, that is, $$\bbox[5px,border:1px solid red] {\color{blue}{kv = kiR}} \tag{2}$$ The additivity property requires that the response to a sum of inputs is the sum of the responses to each input applied separately. Using the voltage-current relationship of a resistor, if $$v_1 = i_1R$$ and $$v_2 = i_2R$$ then applying ($i_1 + i_2$) gives $$v = (i_1 + i_2)R = i_1R + i_2R = v_1 + v_2$$ We say that a resistor is a linear element because the voltage-current relationship satisfies both the homogeneity and the additivity properties.
In general, a circuit is linear if it is both additive and homogeneous.
A linear circuit consists of only linear elements, linear dependent sources, and independent sources. Fig.1: Alinear circuit with input vs and output i.
To understand the linearity principle, consider the linear circuit shown in Fig.1. The linear circuit has no independent sources inside it. It is excited by a voltage source vs , which serves as the input. The circuit is terminated by a load R. We may take the current i through R as the output. Suppose $v_s = 10 V$ gives $i = 2 A$. According to the linearity principle, $v_s = 1 V$ will give $i = 0.2 A$. By the same token, $i = 1 mA$ must be due to $v_s =5 mV$.
Example 1: For the circuit in Fig. 2, find $i_o$ when $v_s = 12 V$ and $v_s = 24 V$. Fig.2:
Solution: Applying KVL to the two loops, we obtain $$12i_1 - 4i_2 + v_s = 0 \tag{1}$$ $$-4i_1 + 16i_2 - 3v_x - v_s = 0\tag{2}$$ But $v_x = 2i_1$. Equation (2) becomes $$-10i_1 + 16i_2 - v_s = 0 \tag{3}$$ Adding Eqs. (1) and (3) yields $$2i_1 + 12i_2 = 0 \Rightarrow i_1 = -6i_2$$ Substituting this in Eq. (1), we get $$-76i_2 + v_s = 0 \Rightarrow i_2 = {v_s \over 76}$$ When vs = 12 V, $$i_o = i_2 = {12 \over 76}A$$ When vs = 24 V, $$i_o = i_2 = {24 \over 76} A$$ showing that when the source value is doubled, $i_o$ doubles.