9.16 Procedures of Solution of Circuit Problems

9.16 Procedures of Solution of Circuit Problems

After solving the above problem, we are in a position to apply the same procedure to analyse other direct current complex networks. While using Kirchhoff's rules in other problems, it is worthwhile to follow the approach given below:

1. Draw the circuit diagram.
2. The choice of loops should be such that each resistance is included at least once in the selected loops.
3. Assume a loop current in each loop. All the loop currents should be in the same sense. It may be either clockwise or anticlockwise.
4. Write the loop equations for all the selected loops. For writing each loop equation, the voltage change across any component is positive if traversed from low to high.

Example 9.11

Calculate the currents in the three resistances of the circuit shown in Fig. 9.30.

Solution:

First we select two loops abcd and ebfc. The choice of loops is quite arbitrary, but it should be such that each resistance is included at least once in the selected loops.

After selecting the loops, suppose a current $I_{1}$ is flowing in the first loop and $I_{2}$ in the second loop, all flowing in the same sense. These currents are called loop currents. The actual currents will be calculated with their help. It should be noted that the sense of the current flowing in all loops should essentially be the same. It may be clockwise or anticlockwise. Here we have assumed it to be clockwise.

We now apply Kirchhoff's second rule to obtain the equations required to calculate the currents through the resistances. We first consider the loop abcda. Starting at point a we follow the loop clockwise. The voltage change while crossing the battery $E_{1}$ is $-E_{1}$, because the current flows through it from positive to negative. The voltage change across $R_{1}$ is $-I_{1}R_{1}$. The resistance $R_{2}$ is common to both the loops $I_{1}$ and $I_{2}$, therefore, the currents $I_{1}$ and $I_{2}$ simultaneously flow through it. The directions of currents $I_{1}$ and $I_{2}$ as flowing through $R_{2}$ are opposite, so we have to decide that which of these currents is to be assigned a positive sign. The convention regarding the sign of the current is that if we are applying the Kirchhoff's second rule in the first loop, then the current of this loop i.e., $I_{1}$, will be assigned a positive sign and all currents flowing opposite to $I_{1}$ have a negative sign. Similarly, while applying Kirchhoff's second rule in the second loop, the current $I_{2}$ will be considered as positive and $I_{1}$ as negative. Using this convention the current flowing through $R_{2}$ is $(I_{1} - I_{2})$ and the voltage change across it is $-(I_{1} - I_{2})R_{2}$. The voltage change across the battery $E_{2}$ is $E_{2}$. Thus, the Kirchhoff's second rule as applied to the loop abcd gives:

Substituting the values, we have:

or

Similarly, applying Kirchhoff's second rule to the loop bcdef, we have:

Substituting the values:

Solving Eq. (i) and Eq. (ii) for $I_{1}$ and $I_{2}$, we have:

Knowing the value of loop currents $I_{1}$ and $I_{2}$, the actual current flowing through each resistance of the circuit can be determined. Fig. 9.29 shows that $I_{1}$ and $I_{2}$ are the actual currents through the resistances $R_{1}$ and $R_{3}$. The actual current through $R_{2}$ is the difference of $I_{1}$ and $I_{2}$ and its direction is along the larger current. Thus,

The current through $R_{1} = I_{1} = \tfrac{2}{3}\,\mathrm{A} = 0.66\,\mathrm{A}$ flowing in the direction of $I_{1}$ i.e., from a to d.

The current through $R_{2} = I_{1} - I_{2} = \tfrac{2}{3}\,\mathrm{A} - \tfrac{2}{9}\,\mathrm{A} = \tfrac{4}{9}\,\mathrm{A} = 0.44\,\mathrm{A}$ flowing in the direction of $I_{1}$ i.e., from c to b.

The current through $R_{3} = I_{2} = \tfrac{2}{9}\,\mathrm{A} = 0.22\,\mathrm{A}$ flowing in the direction of $I_{2}$ i.e., from f to e.