Question

Apply the loop rule to loop akledcba in Figure 21.52.

Short answer

The loop rule to loop abcdefghija is:

$\left(25.1\mathrm{\Omega }\right){\mathrm{I}}_1-\left(40.5\mathrm{\Omega }\right){\mathrm{I}}_2=-24\mathrm{V}$

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

$$\begin{gathered} {\mathrm{R}}_1=5.0\mathrm{\Omega },{\mathrm{E}}_1=24\mathrm{V},{\mathrm{r}}_1=0.10\mathrm{\Omega } \\ {\mathrm{R}}_2=40\mathrm{\Omega },{\mathrm{E}}_2=48\mathrm{V},{\mathrm{r}}_2=0.50\mathrm{\Omega } \\ {\mathrm{R}}_5=20\mathrm{\Omega } \end{gathered}$$

Explanation for the loop akledcba

We go around the loop counter-clockwise, which is the same direction as the current ${\mathrm{I}}_2$but opposite the direction of ${\mathrm{I}}_1$In the direction of decreasing potential, we go through the emf ${\mathrm{E}}_1$.

Therefore, we have

$$\begin{gathered} {\mathrm{E}}_2-{\mathrm{I}}_2{\mathrm{r}}_2-{\mathrm{I}}_2{\mathrm{R}}_2+{\mathrm{I}}_1{\mathrm{R}}_5+{\mathrm{I}}_1{\mathrm{r}}_2-{\mathrm{E}}_2+{\mathrm{I}}_1{\mathrm{R}}_1=0 \\ {\mathrm{E}}_2-{\mathrm{E}}_1-{\mathrm{I}}_1\left({\mathrm{r}}_1+{\mathrm{R}}_1+{\mathrm{R}}_5\right)-{\mathrm{I}}_2\left({\mathrm{r}}_2+{\mathrm{R}}_2\right)=0 \\ \left(25.1\mathrm{\Omega }\right){\mathrm{I}}_1-\left(40.5\mathrm{\Omega }\right){\mathrm{I}}_2=-24\mathrm{V} \end{gathered}$$

Hence, only one loop of the entire circuit is subjected to loop analysis, giving the following expression:

$\left(25.1\mathrm{\Omega }\right){\mathrm{I}}_1-\left(40.5\mathrm{\Omega }\right){\mathrm{I}}_2=-24\mathrm{V}$