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Q29P

Expert-verifiedFound in: Page 797

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?**

- The size of current ${i}_{1}$ is $0.333\text{\hspace{0.17em}A}$
- The direction of current ${i}_{1}$ is rightward
- The energy dissipated by all four resistors in $1.0\text{\hspace{0.17em}min}$ is $720\text{\hspace{0.17em}J}$

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

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

**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:

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

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

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

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{'}}=12.0\Omega $

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,

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

** **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.

Energy dissipated:

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

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

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