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.

##### Linearity Property Related Questions