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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Apply the loop rule to loop abcdefgha in Figure 21.25

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

${\mathrm{l}}_{1}=18.5\mathrm{A}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{2}=2\mathrm{A}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{3}=16.5\mathrm{A}$
See the step by step solution

## Step 1: 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.

## Step 2: 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

$-{\mathrm{E}}_{2}+{\mathrm{l}}_{3}\left({\mathrm{r}}_{2}\right)+\left({\mathrm{l}}_{3}\right)\left({\mathrm{R}}_{3}\right)+\left({\mathrm{l}}_{1}-{\mathrm{l}}_{3}\right){\mathrm{R}}_{1}=0$

## Step 3: 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

${\mathrm{l}}_{2}{\mathrm{R}}_{2}-{\mathrm{E}}_{1}+{\mathrm{l}}_{2}{\mathrm{r}}_{1}+\left({\mathrm{l}}_{1}-{\mathrm{l}}_{3}\right){\mathrm{R}}_{1}=0\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{2}\left(2.5\right)-18+0.5\left({\mathrm{l}}_{2}\right)+6{\mathrm{l}}_{2}=0\phantom{\rule{0ex}{0ex}}\mathrm{Here}\left({\mathrm{l}}_{1}-{\mathrm{l}}_{3}={\mathrm{l}}_{2}\right)\phantom{\rule{0ex}{0ex}}9{\mathrm{l}}_{2}=18\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{2}=2\mathrm{A}$

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:

$2{\mathrm{l}}_{3}+6{\mathrm{l}}_{2}=45\phantom{\rule{0ex}{0ex}}2{\mathrm{l}}_{3}=45-12\phantom{\rule{0ex}{0ex}}2{\mathrm{l}}_{3}=33\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{3}=16.5\mathrm{A}$

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,

${\mathrm{l}}_{2}{\mathrm{R}}_{2}-{\mathrm{E}}_{1}+{\mathrm{l}}_{2}{\mathrm{r}}_{1}+\left({\mathrm{l}}_{1}-{\mathrm{l}}_{3}\right)\mathrm{R}=0\phantom{\rule{0ex}{0ex}}5-18+1+{\mathrm{l}}_{1}\left(6\right)-99=0\phantom{\rule{0ex}{0ex}}-111+6{\mathrm{l}}_{1}=0\phantom{\rule{0ex}{0ex}}6{\mathrm{l}}_{1}=111\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{1}=18.5\mathrm{A}$

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

${\mathrm{l}}_{1}=18.5\mathrm{A}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{2}=2\mathrm{A}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{3}=16.5\mathrm{A}$