Verify the third equation in Example 21.5 by substituting the values found for the currents${\mathbf{l}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{l}}_{\mathbf{3}}$.

Kirchhoff's first law defines currents at circuit junctions. It states that the sum of currents flowing into and out of a junction in an electrical circuit equals the sum of currents flowing out of the junction.

$$\begin{gathered} {\mathrm{E}}_1=18\mathrm{V},{\mathrm{R}}_1=6{\mathrm{\Omega r}}_1=0.5\mathrm{\Omega } \\ {\mathrm{E}}_2=45\mathrm{V},{\mathrm{R}}_2=2.5{\mathrm{\Omega r}}_2=0.5\mathrm{\Omega } \\ {\mathrm{R}}_3=1.5\mathrm{\Omega } \\ {\mathrm{l}}_1=18.5\mathrm{A} \\ {\mathrm{l}}_2=2\mathrm{A} \\ {\mathrm{l}}_3=16.5\mathrm{A} \end{gathered}$$

Applying Kirchhoff's loop rule we obtain:

$-{\mathrm{E}}_2+{\mathrm{l}}_3\left({\mathrm{r}}_2\right)+{\mathrm{l}}_3\left({\mathrm{R}}_3\right)+\left({\mathrm{l}}_1-{\mathrm{l}}_3\right){\mathrm{R}}_1=0$

Consider the equation:

$-{\mathrm{E}}_2+{\mathrm{l}}_3\left({\mathrm{r}}_2\right)+{\mathrm{l}}_3\left({\mathrm{R}}_3\right)+\left({\mathrm{l}}_1-{\mathrm{l}}_3\right){\mathrm{R}}_1=0$

Substituting these values gives zero, showing that the equation is correct.

The sum of all currents entering a location equals the sum of currents leaving the exact point, according to Kirchhoff's junction laws. In addition, the second rule, known as the loop rule, stipulates that the sum of all potential differences in a circuit must equal zero.

Hence, we find that result is equal to zero.