3.13 KIRCHHOFF’S RULES

Electric circuits generally consist of a number of resistors and cells interconnected sometimes in a complicated way. The formulae we have derived earlier for series and parallel combinations of resistors are not always sufficient to determine all the currents and potential differences in the circuit. Two rules, called Kirchhoff’s rules, are very useful for analysis of electric circuits.

Given a circuit, we start by labelling currents in each resistor by a symbol, say $I$ , and a directed arrow to indicate that a current $I$ lows along the resistor in the direction indicated. If ultimately $I$ is determined to be positive, the actual current in the resistor is in the direction of the arrow. If $I$ turns out to be negative, the current actually flows in a direction opposite to the arrow. Similarly, for each source (i.e., cell or some other source of electrical power) the positive and negative electrodes are labelled, as well as, a directed arrow with a symbol for the current flowing through the cell. This will tell us the potential difference, $V = V (P) - V (N) = ε - I r$ [Eq. (3.57) between the positive terminal P and the negative terminal $N$ ; $I$ here is the current flowing from $N$ to $P$ through the cell]. If, while labelling the current $I$ through the cell one goes from $P$ to $N$ , then of course

$V = ε + I r$ (3.79)

3.22

At junction a the current leaving is

I_{1} + I_{2}

and current entering is

I_{3}

. The junction rule says

I_{3} = I_{1} + I_{2}

. At point

h

current entering is

I_{1}

. There is only one current leaving

h

and by junction rule that will also be

I_{1}

. For the loops ‘ahdcba’ and ‘ahdefga’, the loop rules give

- 30 I_{1} - 41 I_{3} + 45 = 0

and

- 30 I_{1} + 21 I_{2} - 80 = 0

Having clarified labelling, we now state the rules and the proof:

Junction rule: At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction (Fig. 3.22).
This applies equally well if instead of a junction of several lines, we consider a point in a line.

The proof of this rule follows from the fact that when currents are steady, there is no accumulation of charges at any junction or at any point in a line. Thus, the total current flowing in, (which is the rate at which charge flows into the junction), must equal the total current flowing out.
Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero (Fig. 3.22).
This rule is also obvious, since electric potential is dependent on the location of the point. Thus starting with any point if we come back to the same point, the total change must be zero. In a closed loop, we do come back to the starting point and hence the rule.

Example 3.6

A battery of $10 V$ and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1 Ω$ (Fig. 3.23). Determine the equivalent resistance of the network and the current along each edge of the cube.

3.23

VIEW SOLUTION

The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.

The paths $A A^{'}$ , $A D$ and $A B$ are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, $I$ . Further, at the corners $A^{'}$ , $B$ and $D$ , the incoming current $I$ must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of $I$ , using Kirchhoff’s first rule and the symmetry in the problem.

Next take a closed loop, say, $A B C C^{'} E A$ , and apply Kirchhoff’s second rule: $- I R - (1 / 2) I R - I R + ε = 0$ where $R$ is the resistance of each edge and $ε$ the emf of battery. Thus, $ε = \frac{5}{2} I R$ The equivalent resistance $R_{e q}$ of the network is $R_{e q} = \frac{ε}{3 I} = \frac{5}{6} R$

For $R = 1 Ω$ , $R_{e q} = (5 / 6) Ω$ and for $ε = 10 V$ , the total current ( $= 3 I$ ) in the network is $3 I = 10 V / (5 / 6) Ω = 12 A, i.e., I = 4 A$ The current flowing in each edge can now be read off from the Fig. 3.23.

It should be noted that because of the symmetry of the network, the great power of Kirchhoff’s rules has not been very apparent in Example 3.6. In a general network, there will be no such simplification due to symmetry, and only by application of Kirchhoff’s rules to junctions and closed loops (as many as necessary to solve the unknowns in the network) can we handle the problem. This will be illustrated in Example 3.7.

Example 3.7

Determine the current in each branch of the network shown in Fig. 3.24.

3.24

VIEW SOLUTION

Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns $I_{1}$ , $I_{2}$ and $I_{3}$ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff’s second rule for the closed loop $A D C A$ gives,

$10 - 4 (I_{1} - I_{2}) + 2 (I_{2} + I_{3} - I_{1}) - I_{1} = 0$ (3.80(a))

that is, $7 I_{1} - 6 I_{2} - 2 I_{3} = 10$

For the closed loop $A B C A$ , we get $10 - 4 I_{2} - 2 (I_{2} + I_{3}) - I_{1} = 0$ that is,

$I_{1} + 6 I_{2} + 2 I_{3} = 10$ (3.80(b))

For the closed loop $B C D E B$ , we get $5 - 2 (I_{2} + I_{3}) - 2 (I_{2} + I_{3} - I_{1}) = 0$ that is,

$2 I_{1} - 4 I_{2} - 4 I_{3} = - 5$ (3.80(c))

Equations (3.80 a, b, c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give $I_{1} = 2.5 A, I_{2} = \frac{5}{8} A, I_{3} = 1 \frac{7}{8} A$

The currents in the various branches of the network are $\begin{array}{l} AB : \frac{5}{8} A, CA : 2 \frac{1}{2} A, DEB : 1 \frac{7}{8} A \\ AD : 1 \frac{7}{8} A, CD : 0 A, BC : 2 \frac{1}{2} A \end{array}$

It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop $B A D E B$ $5 V + (\frac{5}{8} \times 4) V - (\frac{15}{8} \times 4) V$ equal to zero, as required by Kirchhoff’s second rule.