Find the current through each of the three resistors of the circuit shown in Fig. 26.61. The emf sources have negligible internal resistance.

Kirchoff’s law state that at any junction the algebraic sum of all currents should be always equal to zero.

The voltage around a loop equals the sum of every voltage drop in the same loop for any closed network and equals zero

The circuit is sketched below. The current through each of the three resistors are

(through$2\Omega$ )$I_2$(through$5\Omega$ ) and $I_2-I_1$(through$4\Omega$ )

By applying the loop rule on loop 1 and loop 2 we can get the value of unknown currents,$I_{2}and{I}_1$ . We can use loop 3 to correlation check.

For loop 1,

$$\begin{gathered} 20V-I_1\times \left(2\Omega \right)-14V+\left(I_2-I_1\right)\times 4\Omega =0 \\ 3I_1-2I_2=3V \end{gathered}$$

For loop 2,

$$\begin{gathered} 36V-I_2\times \left(5\Omega \right)-\left(I_2-I_1\right)\times 4\Omega =0 \\ -4I_1+9I_2=36V \end{gathered}$$

From loop 1 and loop 2, we get

$$\begin{gathered} I_2=6.32A \\ {I}_1=5.21A \\ ∆I=I_2-I_1=1.11A \end{gathered}$$

Now check with loop 3, we get,

$$\begin{gathered} 20V-I_1\times \left(2\Omega \right)-14V+36V-I_2\times 5\Omega =0 \\ {I}_1\times \left(2\Omega \right)+I_2\times 5\Omega =42A \end{gathered}$$

This expression is verified and the left-hand side is equal to the right-hand side.

Hence, the current in each branch is

$$\begin{gathered} I_2=6.32A \\ {I}_1=5.21A \\ ∆I=I_2-I_1=1.11A \end{gathered}$$