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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Figure ,${{\mathbf{R}}}_{1}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{V}}$ , ${{\mathbf{R}}}_{2}{\mathbf{=}}{\mathbf{18}}{\mathbf{.}}{\mathbf{0}}{\text{\hspace{0.17em}}}{\mathbf{V}}$and the ideal battery has emf ${\mathbf{\epsilon }}{\mathbf{=}}{\mathbf{12}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{V}}$. (a) What is the size of current ${\mathbf{}}{{\mathbf{i}}}_{1}$?(b) What is the direction (left or right) of current ${\mathbf{}}{{\mathbf{i}}}_{1}$ ? (c) How much energy is dissipated by all four resistors in 1.00 min?

1. The size of current ${i}_{1}$ is $0.333\text{\hspace{0.17em}A}$
2. The direction of current ${i}_{1}$ is rightward
3. The energy dissipated by all four resistors in $1.0\text{\hspace{0.17em}min}$ is $720\text{\hspace{0.17em}J}$
See the step by step solution

## Step 1: Write the given data:

${R}_{2}=18.0\text{\hspace{0.17em}}\Omega \phantom{\rule{0ex}{0ex}}{R}_{1}=6.00\text{\hspace{0.17em}}\Omega \phantom{\rule{0ex}{0ex}}V=12.0\text{\hspace{0.17em}V}$

Time, $t=1.00\text{\hspace{0.17em}}\mathrm{min}=60.0\text{\hspace{0.17em}s}$

## Step 2: Understanding the concept

Use the concept of series as well as parallel resistances. Using that, we can find total resistance. Then, we have to use Ohm’s law ${\mathbf{V}}{\mathbf{=}}{\mathbf{IR}}$ to find total current.

Write the formula for series and the parallel resistance:

${R}_{series}={R}_{1}+{R}_{2}\phantom{\rule{0ex}{0ex}}{R}_{parallel}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$

Write the formula for the electrical energy:

$E={i}_{1}^{2}{R}_{1}t$

## Step 3: (a) Calculate the size of current

Here all ${R}_{2}$are connected in parallel. So, the equivalent of those is R .

$\frac{1}{R}=\frac{1}{{R}_{2}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{1}{R}=\frac{1}{18}+\frac{1}{18}+\frac{1}{18}\phantom{\rule{0ex}{0ex}}R=6.00Ω\phantom{\rule{0ex}{0ex}}$

Now ${R}_{1}$ and R are in series, so, equivalent of those is $R\text{'}$.

${R}^{\text{'}}={R}_{1}+R\phantom{\rule{0ex}{0ex}}{R}^{\text{'}}=6.00+6.00\phantom{\rule{0ex}{0ex}}{R}^{\text{'}}=12.0Ω$

Now, according to Ohm’s law, total current l is as follow:

$I=\frac{V}{R\text{'}}\phantom{\rule{0ex}{0ex}}I=\frac{12.0}{12.0}\phantom{\rule{0ex}{0ex}}I=1.00A\phantom{\rule{0ex}{0ex}}$

Now,current ${i}_{1}$ , here I is the total current through $RandR\text{'}$ and R is the combination of three parallel resistances ${R}_{2}.$So, the current through each ${R}_{2}.$ is one third that of the total current.

So,

$\begin{array}{c}{i}_{1}=\frac{I}{3}\\ =\frac{1.00}{3}\\ =0.333\text{\hspace{0.17em}A}\end{array}$

## Step 5: (b) Calculate the direction (left or right) of current i1

Direction of current:

Consider the positive current direction from thepositive terminal to the negative terminal. So, the current l is clockwise means it is rightwards.

## Step 5: (c) Calculate how much energy is dissipated by all four resistors in 1.00 min

Energy dissipated:

$E={I}^{2}R\text{'}t$ Substitute the values and solve for the energy dissipated as:

$E={1.00}^{2}×12.0×60.0\phantom{\rule{0ex}{0ex}}E=720\text{\hspace{0.17em}J}$