Linearity Property


What is a linear circuit?

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,
$$\bbox[5px,border:1px solid grey] {kv = kiR} \tag{2}$$
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
$$v_1 = i_1R$$
$$v_2 = i_2R$$
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.
Alinear circuit with input vs and
output i.
Fig. 1: A linear 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 $v_s$ , 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.

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