Question

Apply the loop rule to loop abcdefgha in Figure 21.25

Short answer

By applying Kirchhoff's loop rule in the loop abcdefgha, we get,

$$\begin{gathered} {\mathrm{l}}_1=18.5\mathrm{A} \\ {\mathrm{l}}_2=2\mathrm{A} \\ {\mathrm{l}}_3=16.5\mathrm{A} \end{gathered}$$

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

$\begin{array}{l}{\mathrm{E}}_1=18\mathrm{V},{\mathrm{R}}_1=6\mathrm{\Omega },{\mathrm{r}}_1=0.5\mathrm{\Omega }\\ {\mathrm{E}}_2=45\mathrm{V},{\mathrm{R}}_2=2.5\mathrm{\Omega },{\mathrm{r}}_2=0.5\mathrm{\Omega }\\ {\mathrm{R}}_3=0.5\mathrm{\Omega }\end{array}$

And

We get the following results using Kirchhoff's loop rule:

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

and

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

Calculation of the current in the loop of abcdegfha

Apply the formula

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

Substituting the values in the above expression, we get

$$\begin{gathered} {\mathrm{l}}_2{\mathrm{R}}_2-{\mathrm{E}}_1+{\mathrm{l}}_2{\mathrm{r}}_1+\left({\mathrm{l}}_1-{\mathrm{l}}_3\right){\mathrm{R}}_1=0 \\ {\mathrm{l}}_2\left(2.5\right)-18+0.5\left({\mathrm{l}}_2\right)+6{\mathrm{l}}_2=0 \\ \mathrm{Here}\left({\mathrm{l}}_1-{\mathrm{l}}_3={\mathrm{l}}_2\right) \\ 9{\mathrm{l}}_2=18 \\ {\mathrm{l}}_2=2\mathrm{A} \end{gathered}$$

Now,

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

Substitute the values in the above expression:

$$\begin{gathered} 2{\mathrm{l}}_3+6{\mathrm{l}}_2=45 \\ 2{\mathrm{l}}_3=45-12 \\ 2{\mathrm{l}}_3=33 \\ {\mathrm{l}}_3=16.5\mathrm{A} \end{gathered}$$

Consider role="math" localid="1655960790613" ${\mathrm{l}}_2{\mathrm{R}}_2-{\mathrm{E}}_1+{\mathrm{l}}_2{\mathrm{r}}_1+\left({\mathrm{l}}_1-{\mathrm{l}}_3\right)\mathrm{R}=0$

Now substitute the value of ${\mathrm{l}}_1$ and ${\mathrm{l}}_3$in the above expression which results in,

$$\begin{gathered} {\mathrm{l}}_2{\mathrm{R}}_2-{\mathrm{E}}_1+{\mathrm{l}}_2{\mathrm{r}}_1+\left({\mathrm{l}}_1-{\mathrm{l}}_3\right)\mathrm{R}=0 \\ 5-18+1+{\mathrm{l}}_1\left(6\right)-99=0 \\ -111+6{\mathrm{l}}_1=0 \\ 6{\mathrm{l}}_1=111 \\ {\mathrm{l}}_1=18.5\mathrm{A} \end{gathered}$$

Hence,after applying the loop rule in loop 'abcdefgha' the values of currents are:

$$\begin{gathered} {\mathrm{l}}_1=18.5\mathrm{A} \\ {\mathrm{l}}_2=2\mathrm{A} \\ {\mathrm{l}}_3=16.5\mathrm{A} \end{gathered}$$