Resistance

Current, Resistance, and EMF

The way viscous friction works, imagine you’re dropping a pebble in oil— you’ve got viscous friction going in the other direction. Theres some coefficient $b$, but the force is proportional to the speed $v$ the thing is moving (whereas air resistance is proportional to $v^2$)

This (viscous friction) is a pretty good analogy from electrons tbh

Current ($I$): the flow of charge from one region to another, defined to be in the direction of a hypothetical flow of positive charges

$$\begin{gathered} I=\frac{dQ}{dt} \\ \frac{[C]}{[S]}=[A] \end{gathered}$$

The current direction is NOT necessarily the same direction in which charged particles are moving. Blame Ben Franklin.

Current Density: $I=\int \vec{J}\cdot d\vec{A}$ $[J]=\frac{A}{m^2}$ $J=\frac{I}{A}$

Thus if a circular wire is tapered so that going from left to right, the width decreases by a factor of two— the current stays the same and the current density quadruples.

Generally in a steady state of flow, the current stays the same. (Just like a river, you cant have more coming in from an area than going out)

Units of resistivity is $\Omega$ ohms

EMF

$V_{\text{terminal}}=\stackrel{EMF}{\overbrace{ϵ}}-I{R}_{\text{internal}}$ $ϵ=I(R_{\text{internal}}+R_{\text{external}})$

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Resistance = Resistivity × Length / (cross-sectional Area) $R=\rho \frac{L}{A}$

Also… $P=VI=\frac{V^2}{R}=I^{2}R$

$I=\frac{EA}{\rho }$

$V=\vec{E}L$ (voltage = electgric field * distance or length or whatever)

$v_d=\frac{I}{n_{\text{electrons}}Aq}$

Resistors in parallel $I=\frac{I}{{\text{R}}_1}+\frac{I}{{\text{R}}_2}+\frac{I}{{\text{R}}_3}$

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Problem 1

Consider two copper wires with the same cross-sectional area. Wire A is twice as long as wire B. How do the resistivities and resistances of the two wires compare?

• Wire A and wire B have the same resistivity.
• Wire A has twice the resistance of wire B.

Problem 4 A current-carrying gold wire has a diameter of 0.80 $\text{mm}$. The electric field in the wire is 0.47 $\text{V}/\text{m}$. Use the resistivity at room temperature for gold $\rho =2.44\times 10^{-8}\Omega \cdot \text{m}$.

What is the current carried by the wire?

$$\begin{array}{rl}& \text{let’s use the equation }I=\frac{EA}{\rho }\text{ where:}\\ & \text{‘A’ is the cross sectional area }\pi (\frac{d}{2}{)}^2\\ & \text{we need our d in SI units, so lets do }0.80\text{ mm}=0.0008\text{ m}\\ & \text{now lets get our area }A=\pi (\frac{0.0008}{2}{)}^2=5.03\times 10^{-7}{\text{ m}}^2\\ & \text{then we can just plug and chug into }I=\frac{EA}{\rho }\text{ yielding...}\\ & =\frac{(0.47\text{ V/m})(5.03\times 10^{-7}{\text{ m}}^2)}{2.44\times 10^{-8}\Omega \cdot \text{m}}\\ & =9.67\text{ A..... omg}\end{array}$$

What is the potential difference between two points in the wire $6.9\text{m}$ apart?

Ok, “potential difference” means volts, Volts $\text{(V)}$! Thus, we can just do $V=\vec{E}L$

$$\begin{array}{rl}V& =\vec{E}L\\ & =(0.47)\frac{V}{\cancel{m}}\cdot 6.9\cancel{m}\\ & =(0.47)(6.9)V\\ & =3.243V\end{array}$$

What is the resistance of a $6.9\text{m}$ length of this wire?

Well since $V=IR$ we can literally just plug and chug like

$$\begin{array}{rl}R& =\frac{V}{I}\\ & =\frac{(3.243V)}{(9.67A)}\\ & =0.33536711478\Omega \end{array}$$

Problem 5

An electrical conductor designed to carry large currents has a circular cross section $2.90$ $\text{mm}$ in diameter and is $13.2$ $\text{m}$ long. The resistance between its ends is $0.102$ $\Omega$.

What is the resistivity of the material?

$$\begin{array}{rl}R& =\rho \frac{L}{A}\\ R\frac{A}{L}& =\rho \\ \rho & =R\frac{A}{L}\\ & =R\frac{\pi (\frac{d}{2}{)}^2}{L}\\ & =(0.102\Omega )\frac{\pi (\frac{2.90\times 10^{-3}m}{2}{)}^2}{13.2\text{ m}}\\ & =(0.102)\frac{\pi (1.45\times 10^{-3}{)}^2}{13.2}\Omega \text{m}\\ & =(0.102)\frac{3.14159(2.1025\times 10^{-6})}{13.2}\Omega \text{m}\\ & =(0.102)\frac{6.6046\times 10^{-6}}{13.2}\Omega \text{m}\\ & =5.1\times 10^{-8}\Omega \text{m}\end{array}$$

If the electric field magnitude in the conductor is $1.25\frac{V}{m}$, what is the total current?

$$\begin{array}{rl}V& =IR\\ I& =\frac{V}{R}\\ I& =\frac{\vec{E}L}{R}\\ & =\frac{(1.25\frac{V}{\cancel{m}})(13.2\cancel{m})}{0.102\Omega }\\ & =\frac{(1.25\ast 13.2)V}{0.102\Omega }\\ & =\frac{16.5\cancel{V}}{0.102\cancel{\Omega }}\\ & =\frac{16.5}{0.102}\text{A}\\ & =161.764705882\text{A}\end{array}$$

If the material has $8.5\times 10^{28}$ free electrons per cubic meter, find the average drift speed under the conditions of the previous part.

$$\begin{array}{rl}{v}_d& =\frac{I}{n_{\text{electrons}}Aq}\\ & =\frac{I}{n_{\text{electrons}}\pi {\left(\frac{d}{2}\right)}^{2}q}\\ & =\frac{161.764705882\text{A}}{(8.5\times 10^{28})(6.6046\times 10^{-6}{\text{m}}^2)(1.60217663\cdot 10^{-19}\text{C})}\\ & =\frac{161.764705882}{(8.5\times 10^{28})(6.6046\times 10^{-6})(1.60217663\cdot 10^{-19})}\cdot \frac{\text{A}}{{\text{m}}^2\text{C}}\\ & =\frac{161.764705882}{(8.5\times 10^{28})(6.6046\times 10^{-6})(1.60217663\cdot 10^{-19})}\cdot \frac{\text{m}}{2}\\ & =\frac{161.764705882}{(8.5)(6.6046)(1.60217663)\times 10^3}\cdot \frac{\text{m}}{2}\\ & =\frac{161.764705882}{(8.5)(6.6046)(1.60217663)}\times 10^{-3}\cdot \frac{\text{m}}{2}\\ & =\frac{161.764705882}{89.9447540492}\times 10^{-3}\cdot \frac{\text{m}}{2}\\ & =1.79848961278\times 10^{-3}\cdot \frac{\text{m}}{2}\end{array}$$

Problem 6

A person with body resistance between his hands of $10\text{k}\Omega$ accidentally grasps the terminals of a $16\text{kV}$ power supply.

Let’s convert out to standard units, idiot grabs a power supply with a voltage of $16\text{kV}$ which is $16000\text{V}$ and his resistance (in his hands) is $10\text{k}\Omega$ which equals $10000\Omega$.

If the internal resistance of the power supply is $2200\Omega$ , what is the current through the person’s body?

$$\begin{array}{rl}V& =IR\\ I& =\frac{V}{R}\\ & =\underset{\text{Resistances add 😀}}{\underbrace{\frac{V}{R_{\text{power suply}}+R_{\text{hands}}}}}\\ & =\frac{16000\text{V}}{2200\Omega +10000\Omega }\\ & =\frac{160\cancel{00}\cancel{\text{V}}}{122\cancel{00}\cancel{\Omega }}\\ & =\frac{160}{122}\text{A}\\ & =1.31147540984\text{A}\\ & \text{This current is lethal.}\end{array}$$

What is the power dissipated in his body?

$$\begin{array}{rl}& \\ {\text{P}}_{\text{body}}& ={\text{I}}^2{\text{R}}_{\text{body}}\\ & =(\stackrel{\text{Lethal current}}{\overbrace{1.31147540984\text{A}}}{)}^2(10000\Omega )\\ & =(1.31147540984{)}^2(10000)\text{W}\\ & =17199.6775061\text{W}\\ & \text{17 kW through human tissue would cause severe burns in addition to the lethal current effects.}\end{array}$$

If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be $I_{\text{max}}=1.0\text{mA}$ or less?

$$\begin{array}{rl}{I}_{\text{max}}& =1.0\text{mA}\\ I_{\text{max}}& =0.001\text{A}\\ I_{\text{max}}& =0.001\text{A}=\frac{\text{V}}{\text{R}}\\ 0.001\text{A}& =\frac{\text{V}}{{\text{R}}_{\text{power supply (internal)}}+{\text{R}}_{\text{hands}}}\\ 0.001\text{A}& =\frac{16000\text{V}}{{\text{R}}_{\text{internal}}+10000\Omega }\\ (0.001\text{A})({\text{R}}_{\text{internal}}+10000\Omega )& =\frac{16000\text{V}}{1}\\ {\text{R}}_{\text{internal}}+10000\Omega & =\frac{16000\cancel{\text{V}}}{0.001\cancel{\text{A}}}\\ {\text{R}}_{\text{internal}}+10000\Omega & =\frac{16000}{0.001}\Omega \\ {\text{R}}_{\text{internal}}+10000\Omega & =16000000\Omega \\ {\text{R}}_{\text{internal}}& =15990000\Omega \end{array}$$

Is he cooked?

• [f] The victim would be “cooked” in multiple ways: There’s the explosive tissue damage from rapid steam formation, severe tissue burns from $17.2\text{kW}$ power dissipation and also an immediate cardiac arrest from the $1.31\text{A}$ current

Problem 7

Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a $100\text{ W}$ incandescent bulb uses only $23.0\text{ W}$ of power. The compact bulb lasts $10000$ hours, on the average, and costs \11.00$,whereastheincandescentbulbcostsonly$$0.76$,butlastsjust$750$hours.Thestudyassumedthatelectricitycost$$0.090$perkilowatt-hourandthatthebulbswereonfor$4.0 \text{ hours}$ per day.

$$\begin{array}{rl}{\text{hours}}_{\text{per year}}& =(365.0)\stackrel{\text{Hours per day}}{\overbrace{(4.0)}}\\ {\text{hours}}_{\text{per year}}& =1460\\ {\text{hours}}_{\text{total}}& =3(1460)\\ & =4380\\ & \\ {\text{price}}_{{\text{per unit}}_{\text{compact}}}& =\underset{\text{Lasts 10K hours}}{\underbrace{\$11.00}}\\ {\text{price}}_{{\text{per unit}}_{\text{incandescent}}}& =\underset{\text{Lasts 750 hours}}{\underbrace{\$0.76}}\\ {\text{price}}_{{\text{total}}_{\text{compact}}}& ={\text{price}}_{{\text{per unit}}_{\text{compact}}}\cdot \frac{{\text{hours}}_{\text{total}}}{10000}\\ {\text{price}}_{{\text{total}}_{\text{incandescent}}}& ={\text{price}}_{{\text{per unit}}_{\text{incandescent}}}\cdot \frac{{\text{hours}}_{\text{total}}}{750}\\ {\text{price}}_{{\text{total}}_{\text{compact}}}& =\$11.00\cdot \frac{4830}{10000}\\ {\text{price}}_{{\text{total}}_{\text{incandescent}}}& =\$0.76\cdot \frac{4380}{750}\\ {\text{price}}_{{\text{total}}_{\text{compact}}}& =\$11.00\cdot \text{ceil}(0.483)\\ {\text{price}}_{{\text{total}}_{\text{incandescent}}}& =\$0.76\cdot \text{ceil}(5.84)\\ {\text{price}}_{{\text{total}}_{\text{compact}}}& =\$11.00\\ {\text{price}}_{{\text{total}}_{\text{incandescent}}}& =\$4.56\\ & \\ & {\text{price}}_{\text{total compact}}=\$11.00\\ & {\text{price}}_{\text{total incandescent}}=\$4.56\\ & {\text{hours}}_{\text{total}}=4380\end{array}$$

What is the total cost (including the price of the bulbs) to run incandescent bulbs for 3.0 years?

$$\begin{array}{rl}{\text{electricity}}_{\text{rate}}& =\$0.090\frac{\text{kWh}}{\text{hour}}\\ {\text{power}}_{\text{incandescent}}& =100\text{ W}\cdot \frac{1\text{ kW}}{1000\text{ W}}=0.1\text{ kW}\\ \text{cost}& =({\text{electricity}}_{\text{rate}}\times {\text{power}}_{\text{incandescent}}\times \text{hours})+{\text{cost}}_{\text{incandescent}}\\ & =((\$0.090\frac{\cancel{\text{kWh}}}{\cancel{\text{hour}}})(0.1\cancel{\text{ kW}})(4380))+\$4.56\\ & =((\$0.090)(0.1)(4380))+\$4.56\\ & =\$39.42+\$4.56\\ & =\$43.98\end{array}$$

What is the total cost (including the price of the bulbs) to run compact fluorescent bulbs for 3.0 years?

$$\begin{array}{rl}{\text{electricity}}_{\text{rate}}& =\$0.090\frac{\text{kWh}}{\text{hour}}\\ {\text{power}}_{\text{compact}}& =23\text{ W}\cdot \frac{1\text{ kW}}{1000\text{ W}}=0.023\text{ kW}\\ \text{cost}& =({\text{electricity}}_{\text{rate}}\times {\text{power}}_{\text{compact}}\times \text{hours})+{\text{cost}}_{\text{compact}}\\ & =((\$0.090\frac{\cancel{\text{kWh}}}{\cancel{\text{hour}}})(0.023\cancel{\text{ kW}})(4380))+\$11.00\\ & =((\$0.090)(0.023)(4380))+\$11.00\\ & =\$9.0666+\$11.00\\ & =\$20.0666\end{array}$$

How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb?

$$\$43.98-\$20.0666=\$23.9134$$

What is the resistance of a "$100\text{W}$" fluorescent bulb? (Remember, it actually uses only $23\text{W}$ of power and operates across $120\text{V}$.)

$$\begin{array}{rl}P& =VI\\ P& =\frac{V^2}{R}\\ R& =\frac{V^2}{P}\\ & =\frac{(120\text{V}{)}^2}{(23\text{W})}\\ & =\frac{120^2{\text{V}}^2}{23\text{W}}=(\frac{120^2}{23})\cancel{\frac{{\text{V}}^2}{\text{W}}}\\ & =\frac{120^2}{23}\Omega \\ & =626.086956522\Omega \end{array}$$

Problem 8

Consider the circuit shown in the following figure. The emf source has negligible internal resistance. The resistors have resistances $R_1=7.00\Omega$ and $R_2=4.00\Omega$. The capacitor has capacitance $C=8.00\mu \text{F}$. When the capacitor is fully charged, the magnitude of the charge on its plates is $Q=32.0\mu \text{C}$. Calculate the emf $ϵ$.

Remember:

• the capacitor acts like an open circuit, the current is not flowing
• the voltage across the capacitor equals the voltage at its terminals

$$\begin{array}{rl}{V}_{\text{capacitor}}& =\frac{Q}{C}\\ & =\frac{32.0\cancel{\mu C}}{8.00\cancel{\mu F}}\\ & =\frac{32.0}{8.00}\text{V}\\ & =4.00\text{V}\\ & \\ V_{\text{capacitor}}& =\stackrel{\text{where tf?}}{\overbrace{\frac{R_1}{R_1+R_2}}}ϵ\\ 4.00\text{V}& \\ & =\frac{7\Omega }{7\Omega +4\Omega }ϵ\\ 4.00\text{V}& \\ & =\frac{7}{11}\Omega ϵ\\ 4.00(\frac{11}{7})\text{V}& \\ & =ϵ\\ 6.29\text{V}& \\ & =ϵ\end{array}$$

Problem 9

An external resistor $R$ is connected between the terminals of a battery. The value of $R$ varies. For each $R$ value, the current $I$ in the circuit and the terminal voltage $V_{ab}$ of the battery are measured. The results are plotted in the figure, a graph of $V_{ab}$ versus $I$ that shows the best straight-line fit to the data.

Use the graph in the figure to calculate the battery’s emf.

Use the graph in the figure to calculate the battery’s internal resistance.

$$\begin{array}{rlr}y=mx+b& & m=\frac{\text{rise}}{\text{run}}\\ y-y_1=m(x-x_1)& & m=\frac{y_2-y_1}{x_2-x_1}\\ & & =\frac{30-24}{3-6}\\ & & =-\frac{6}{3}\\ y-30& =-2(x-3.00)& \\ & & \\ y-30& =-2x+6.00& \\ & & \\ V_{\text{terminal}}& =ϵ-R_{\text{internal}}& \\ V_{\text{terminal}}& =-R_{\text{internal}}+ϵ& \\ & & \\ y& =-2x+36.00& \\ (V_{ab}{)}_y& =\underset{{\text{R}}_{\text{internal}}}{\underbrace{-2\Omega }}{I}_x\text{A}+\underset{ϵ}{\underbrace{36.00\text{V}}}& \end{array}$$

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