2.4 Kirchhoff's Laws

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2.4 Kirchhoff’s Laws

Ohm’s law by itself is not sufficient to analyze circuits. However, when it is coupled with Kirchhoff’s two laws, we have a sufficient, powerful set of tools for analyzing a large variety of electric circuits. Kirchhoff’s laws were first introduced in 1847 by the German physicist Gustav Robert Kirchhoff (1824–1887). These laws are formally kno wn as Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL).

Kirchhoff’s first law is based on the la w of conservation of charge, which requires that the algebraic sum of charges within a system cannot change.

Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero.

Mathematically, KCL implies that

$$\sum_{n=1}^{N} i_n = 0$$

(2.13)

where N is the number of branches connected to the node and in is the nth current entering (or lea ving) the node. By this la w, currents entering a node may be regarded as positive, while currents leaving the node may be taken as negative or vice versa.

Historical

Gustav Robert Kirchhoff (1824–1887), a German physicist, stated two basic laws in 1847 concerning the relationship between the cur rents and voltages in an electrical network. Kirchhoff’s laws, along with Ohm’s law, form the basis of circuit theory.

Born the son of a lawyer in Konigsberg, East Prussia, Kirchhoff entered the University of Konigsberg at age 18 and later became a lecturer in Berlin. His collaborative work in spectroscopy with German chemist Robert Bunsen led to the discovery of cesium in 1860 and rubidium in 1861. Kirchhoff was also credited with the Kirchhoff law of radiation. Thus, Kirchhoff is famous among engineers, chemists, and physicists.

To prove KCL, assume a set of currents i ^k (t) , k = 1, 2,…, flow into a node. The algebraic sum of currents at the node is

$$i_T(t) = i_1(t) + i_2(t) + i_3(t) + \cdots$$

(2.14)

Integrating both sides of Eq. (2.14) gives

$$q_T(t) = q_1(t) + q_2(t) + q_3(t) + \cdots$$

(2.15)

where qk (t) = ∫ ik (t) d t and qT (t) = ∫ iT (t) d t .But the law of conservation of electric charge requires that the algebraic sum of electric charges at the node must not change; that is, the node stores no net charge. Thus, qT (t) = 0 → iT (t) = 0, confirming the validity of KCL.

Consider the node in Fig. 2.16. Applying KCL gives

$$i_1 + (-i_2) + i_3 + i_4 + (-i_5) = 0 \tag{2.16}$$

since currents i1, i3, and i4 are entering the node, while currents i2 and i⁵ are leaving it. By rearranging the terms, we get

$$i_1 + i_3 + i_4 = i_2 + i_5 \tag{2.17}$$

Equation (2.17) is an alternative form of KCL:

The sum of the currents entering a node is equal to the sum of the currents leaving the node.

Note that KCL also applies to a closed boundary . This may be re garded as a generalized case, because a node may be regarded as a closed surface shrunk to a point. In tw o dimensions, a closed boundary is the same as a closed path. As typically illustrated in the circuit of Fig. 2.17, the total current entering the closed surface is equal to the total current leaving the surface.

A simple application of KCL is combining current sources in parallel. The combined current is the algebraic sum of the current supplied by the indi vidual sources. F or example, the current sources sho wn in

Figure 2.16 Currents at a node illustrating KCL.

Figure 2.17 Applying KCL to a closed boundary.

Two sources (or circuits in general) are said to be equivalent if they have the same i-v relationship at a pair of terminals.

Fig. 2.18(a) can be combined as in Fig. 2.18(b). The combined or equivalent current source can be found by applying KCL to node a.

$$I_T + I_2 = I_1 + I_3$$

or

$$I_T = I_1 - I_2 + I_3 \tag{2.18}$$

A circuit cannot contain two different currents, I1 and I2, in series, unless I1 =I2; otherwise KCL will be violated.

Kirchhoff’s second law is based on the principle of conservation of energy:

Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.

Expressed mathematically, KVL states that

$$\sum_{m=1}^{M} v_m = 0$$

(2.19)

where M is the number of voltages in the loop (or the number of branches in the loop) and vm is the mth voltage.

To illustrate KVL, consider the circuit in Fig. 2.19. The sign on each voltage is the polarity of the terminal encountered first as we travel around the loop. We can start with an y branch and go around the loop either clockwise or counterclockwise. Suppose we start with the v oltage source and go clockwise around the loop as sho wn; then v oltages would be −v1, +v2, +v3, +v4, and −v5, in that order. For example, as we reach branch 3, the positive terminal is met first; hence, we have +v3. For branch 4, we reach the ne gative terminal first; hence, −v4. Thus, KVL yields

$$-v_1 + v_2 + v_3 - v_4 + v_5 = 0 \tag{2.20}$$

Rearranging terms gives

$$v_2 + v_3 + v_5 = v_1 + v_4 \tag{2.21}$$

which may be interpreted as

Sum of voltage drops = Sum of voltage rises	(2.22)
---------------------------------------------	--------

This is an alternative form of KVL. Notice that if we had traveled counterclockwise, the result w ould have been +v1, −v5, +v4, −v3, and −v2, which is the same as before e xcept that the signs are re versed. Hence, Eqs. (2.20) and (2.21) remain the same.

When voltage sources are connected in series, KVL can be applied to obtain the total v oltage. The combined v oltage is the algebraic sum of the v oltages of the indi vidual sources. F or example, for the v oltage sources shown in Fig. 2.20(a), the combined or equivalent voltage source in Fig. 2.20(b) is obtained by applying KVL.

$$-V_{ab} + V_1 + V_2 - V_3 = 0$$

Figure 2.18 Current sources in parallel: (a) original circuit, (b) equivalent circuit.

KVL can be applied in two ways: by taking either a clockwise or a counterclockwise trip around the loop. Either way, the algebraic sum of voltages around the loop is zero.

Figure 2.19 A single-loop circuit illustrating KVL.

$$V_{ab} = V_1 + V_2 - V_3 \tag{2.23}$$

To avoid violating KVL, a circuit cannot contain tw o different voltages V1 and V2 in parallel unless V1 =V2.

Figure 2.20

Voltage sources in series: (a) original circuit, (b) equivalent circuit.

^For the circuit in Fig. 2.21(a), find voltages v1 and v² Example 2.5 .

Solution:

To find v1 and v2 we apply Ohm’s law and Kirchhoff’s voltage law. Assume that current i flows through the loop as shown in Fig. 2.21(b). From Ohm’s law,

$$v_1 = 2i, \qquad v_2 = -3i \tag{2.5.1}$$

Applying KVL around the loop gives

$$-20 + v_1 - v_2 = 0 \tag{2.5.2}$$

Substituting Eq. (2.5.1) into Eq. (2.5.2), we obtain

$$-20 + 2i + 3i = 0 \qquad \text{or} \qquad 5i = 20 \qquad \Rightarrow \qquad i = 4 \text{ A}$$

Substituting i in Eq. (2.5.1) finally gives

$$v_1 = 8 \text{ V}, \qquad v_2 = -12 \text{ V}$$

Answer: 16 V, −8 V.

Determine ^v Example 2.6 ^o and i in the circuit shown in Fig. 2.23(a).

Figure 2.23

For Example 2.6.

Solution:

We apply KVL around the loop as shown in Fig. 2.23(b). The result is

$$-12 + 4i + 2v_o - 4 + 6i = 0 \tag{2.6.1}$$

Applying Ohm’s law to the 6-Ω resistor gives

$$v_o = -6i \tag{2.6.2}$$

Substituting Eq. (2.6.2) into Eq. (2.6.1) yields

$$-16 + 10i - 12i = 0 \qquad \Rightarrow \qquad i = -8 \text{ A}$$

and vo = 48 V.

Answer: 20 V, −10 V.

For Example 2.7.

Find vo and io in the circuit of Fig. 2.26.

Practice Problem 2.7

Example 2.8 Find currents and voltages in the circuit shown in Fig. 2.27(a).

For Example 2.8.

Solution:

We apply Ohm’s law and Kirchhoff’s laws. By Ohm’s law,

$$v_1 = 8i_1, \qquad v_2 = 3i_2, \qquad v_3 = 6i_3 \tag{2.8.1}$$

Since the voltage and current of each resistor are related by Ohm’s law as shown, we are really looking for three things: (v1, v2, v3) or (i1, i2, i3). At node a, KCL gives

$$i_1 - i_2 - i_3 = 0 \tag{2.8.2}$$

Applying KVL to loop 1 as in Fig. 2.27(b),

$$-30 + v_1 + v_2 = 0$$

We express this in terms of i1 and i2 as in Eq. (2.8.1) to obtain

$$-30 + 8i_1 + 3i_2 = 0$$

or

$$i_1 = \frac{(30 - 3i_2)}{8} \tag{2.8.3}$$

Applying KVL to loop 2,

−v² + v³ = 0 ⇒ v³ = v² (2.8.4)

as expected since the two resistors are in parallel. We express v1 and v2 in terms of i1 and i2 as in Eq. (2.8.1). Equation (2.8.4) becomes

$$6i_3 = 3i_2$$

$\Rightarrow$ $i_3 = \frac{i_2}{2}$ (2.8.5)

Substituting Eqs. (2.8.3) and (2.8.5) into (2.8.2) gives

$$\frac{30 - 3i_2}{8} - i_2 - \frac{i_2}{2} = 0$$

or i2 = 2 A. From the v alue of i2, we now use Eqs. (2.8.1) to (2.8.5) to obtain

$$i_1 = 3$$

A, $i_3 = 1$ A, $v_1 = 24$ V, $v_2 = 6$ V, $v_3 = 6$ V

Find the currents and voltages in the circuit shown in Fig. 2.28. Practice Problem 2.8

Answer:

$$v_1 = 6$$

V, $v_2 = 4$ V, $v_3 = 10$ V, $i_1 = 3$ A, $i_2 = 500$ mA, $i_3 = 2.5$ A.

2.5 Series Resistors and Voltage Division

The need to combine resistors in series or in parallel occurs so frequently that it warrants special attention. The process of combining the resistors is facilitated by combining two of them at a time. With this in mind, consider the single-loop circuit of Fig. 2.29. The two resistors are in series, since the same current i flows in both of them. Applying Ohm’s law to each of the resistors, we obtain

$$v_1 = iR_1, \qquad v_2 = iR_2 \tag{2.24}$$

If we apply KVL to the loop (mo ving in the clockwise direction), we have

$$-v + v_1 + v_2 = 0 \tag{2.25}$$

Combining Eqs. (2.24) and (2.25), we get

$$v = v_1 + v_2 = i(R_1 + R_2)$$

(2.26)

v + – R1 v1 R2 v2 i + – + – a b

Figure 2.29 A single-loop circuit with two resistors in series.

or

$$i = \frac{v}{R_1 + R_2} \tag{2.27}$$

Figure 2.28 For Practice Prob. 2.8.

Notice that Eq. (2.26) can be written as

$$v = iR_{\text{eq}} \tag{2.28}$$

implying that the two resistors can be replaced by an equivalent resistor Req; that is,

$$R_{\text{eq}} = R_1 + R_2 \tag{2.29}$$

Thus, Fig. 2.29 can be replaced by the equivalent circuit in Fig. 2.30. The two circuits in Figs. 2.29 and 2.30 are equivalent because they exhibit the same voltage-current relationships at the terminals a-b. An equivalent circuit such as the one in Fig. 2.30 is useful in simplifying the analysis of a circuit. In general,

The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.

For N resistors in series then,

$$R_{\text{eq}} = R_1 + R_2 + \dots + R_N = \sum_{n=1}^{N} R_n \tag{2.30}$$

To determine the voltage across each resistor in Fig. 2.29, we substitute Eq. (2.26) into Eq. (2.24) and obtain

$$v_1 = \frac{R_1}{R_1 + R_2} v, \qquad v_2 = \frac{R_2}{R_1 + R_2} v$$

(2.31)

Notice that the source v oltage v is divided among the resistors in direct proportion to their resistances; the larger the resistance, the larger the voltage drop. This is called the principle of voltage division, and the circuit in Fig. 2.29 is called a voltage divider. In general, if a voltage divider has N resistors (R1, R2, … , RN) in series with the source voltage v, the nth resistor (Rn) will have a voltage drop of

… ,

$$R_N$$

) in series with the source voltage v, the nth
a voltage drop of

$$v_n = \frac{R_n}{R_1 + R_2 + \dots + R_N} v$$

(2.32)

2.6 Parallel Resistors and Current Division

Consider the circuit in Fig. 2.31, where tw o resistors are connected in parallel and therefore ha ve the same v oltage across them. From Ohm’s law,

$$v = i_1 R_1 = i_2 R_2$$ $$i_1 = \frac{v}{R_1}$$

, $i_2 = \frac{v}{R_2}$ (2.33)

Applying KCL at node a gives the total current i as

$$i = i_1 + i_2 \tag{2.34}$$

Substituting Eq. (2.33) into Eq. (2.34), we get

or

$$i = \frac{v}{R_1} + \frac{v}{R_2} = v\left(\frac{1}{R_1} + \frac{1}{R_2}\right) = \frac{v}{R_{\text{eq}}}$$

(2.35)

Figure 2.31 Two resistors in parallel.

Equivalent circuit of the Fig. 2.29 circuit.

Resistors in series behave as a single resistor whose resistance is equal to the sum of the resistances of the

individual resistors.