Question

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

Short answer

The sum of all potential differences in a circuit must equal zero.

Step-by-step solution

Definition of Kirchhoff's law

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.

Given information and formula to be used

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

Applying Kirchhoff's loop rule we obtain:

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

Explanation

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.