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Found in: Page 766

### College Physics (Urone)

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

# Verify the second equation in Example 21.5 by substituting the values found for the currents ${{\mathbf{l}}}_{{\mathbf{1}}}$ and ${{\mathbf{l}}}_{{\mathbf{2}}}$

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

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

${\mathrm{E}}_{1}=18\mathrm{V},{\mathrm{R}}_{1}=6{\mathrm{\Omega r}}_{1}=0.5\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{2}=45\mathrm{V},{\mathrm{R}}_{2}=2.5{\mathrm{\Omega r}}_{2}=0.5\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{3}=1.5\mathrm{\Omega }$

Applying Kirchhoff's loop rule we obtain:

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

## Step 3: To verify

Consider the equation:

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

Substitute the values:

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

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.