Linearity Property

Electrical Circuit Analysis > Network Theorems

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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

$$v_1 = i_1R$$

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: 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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