a. Resistance is an opposition to the flow of electrical current.

b. $mil=\frac{1}{1000}$ of a unit. In sizing wires, a mil represents the diameter of the wire in thousandths of an inch.

c. Circular Mils represent the unit of measurement used for describing the cross-sectional area of a wire.

d. A mil-foot is the unit of resistivity of different metals. Its number represents the number of ohms exhibited by a one foot long wire, having a diameter of one mil.

The resistance of a wire:

• is proportional to its length.
• is inversely proportional to its thickness.
• depends on the material from which it is made.
• changes with variations of temperature.

Nickel-chromium, (Nichrome for short) is generally used for the construction of heating elements.

A small resistance value causes a large current and therefore, plenty of heat.

Note: Beginning students often equate resistance with heat, expressing their mistaken belief that lots of ohms produce lots of heat.

To point out the fallacy of such thinking, the instructor might do well to remind the student that air has an infinitely high resistance and, by such reasoning, would make a marvelous space heater.

The 20-foot-long wire has more resistance. The longer the wire, the larger its resistance.

A #14 wire is smaller in diameter than a #12 wire. Therefore, the #14 wire has more resistance for any given length.

As the lamp is turned on, the resistance of the filament increases sharply.

Note: This can well be demonstrated as follows:

With an ohmmeter, measure the resistance of an ordinary 100 W light bulb. It should measure approximately 10 ohms.

Next, measure the current through the lamp as it is connected to a 120 V supply. The current should be approximately 0.8 amperes. Use Ohm’s Law to demonstrate that the resistance has increased from approximately 10 ohms to approximately 150 ohms.

$$R=\frac{\textrm{120 V}}{\textrm{0.8 A}}$$

A hot lamp draws less current than a cold lamp because the resistance of its filament increases significantly as its temperature rises. When the lamp is first switched on, the cold filament has low resistance and draws a large inrush current. As the filament heats up during normal operation, its resistance increases, reducing the current to its rated operating value.

Neither wire has more resistance. Since 17 feet is equal to 204 inches, both wires have the same length. Because they are made from the same material and have the same cross-sectional area, they have equal electrical resistance.

$$A_{\mathrm{CM}}=(d_{\mathrm{mil}})^2=(64)^2=4096\ \mathrm{CM}$$

a. $$0.012\,\text{in.}=12\,\text{mils}$$

$$A_{\mathrm{CM}}=(d_{\mathrm{mil}})^2=(12)^2=144\ \mathrm{CM}$$

b. $$0.0155\,\text{in.}=15.5\,\text{mils}$$

$$A_{\mathrm{CM}}=(15.5)^2=240\ \mathrm{CM}$$

$$d_{\mathrm{mil}}=\sqrt{\mathrm{CM}}=\sqrt{81}=9\,\mathrm{mil}$$

$$9\,\text{mils}=0.009\,\text{inches}$$

The given equation $R=\frac{K\times L}{CM}$ demands that the length L be stated in feet.

a.     $$R=\frac{KL}{\mathrm{CM}}=\frac{17\times100}{4107}=0.414\,\Omega$$

b.    $$R=\frac{KL}{\mathrm{CM}}=\frac{600\times25}{1022}=14.7\,\Omega$$

c.     $$R=\frac{KL}{\mathrm{CM}}=\frac{60\times5280}{16510}=19.2\,\Omega$$

d.    $$R=\frac{KL}{\mathrm{CM}}=\frac{10.4\times0.5}{1624}=0.0032\,\Omega$$

a. To find the copper wire size to meet these requirements, either one of two methods can be used.

$$R=\frac{KL}{\mathrm{CM}}$$

$$5=\frac{10.4\times12000}{\mathrm{CM}}$$

$$5(\mathrm{CM})=124800$$

$$\mathrm{CM}=\frac{124800}{5}=24960\ \mathrm{CM}$$

Thus, No. 6 wire can be used (26,250 CM)

b. Use $$R=\frac{KL}{CM}$$ to find the smallest size of aluminum wire that meets the requirements

$$5=\frac{17\times 12,000}{CM}$$

$$CM=41,740 \left ( \textrm{this value requires No. 4 wire} \right )$$

A DC-generator winding has a resistance of 54.5 ohms at 20°C. The resistance of the copper winding after a temperature increase to 50°C is determined as follows:

$$\text{Resistance Increase}=R\times\text{coefficient}\times\text{degree rise}$$

$$=54.5\times0.004\times30$$

$$=6.54\,\Omega$$

Thus, the resistance at 50 = 54.5 + 6.54 = 61 ohms.

A coil of copper wire has a resistance of 150 ohms at 20°C. After several hours of operation, the wire resistance is 172 ohms. The temperature of the wire is determined as follows:

$$\text{Degree Rise}=\frac{\text{Resistance Increase}}{R\times\text{coefficient}}=\frac{22}{150\times0.004}=36.66^\circ$$

$$\therefore\ \text{Degree Rise}=20+36.66=56.66^\circ\text{C}\approx56.7^\circ\text{C}$$

$$d_{\mathrm{mil}}=\sqrt{\mathrm{CM}}=\sqrt{4096}=64\,\text{mils}=0.064\,\text{in.}$$

$$\text{Each strand has a C.S.A.}=(8.75)^2=76.56\ \mathrm{CM}$$

$$\text{The area of 74 strands}=74\times76.56=5665\ \mathrm{CM}$$

$$19\text{ strands have }1200\,\mathrm{MCM}=1200000\,\mathrm{CM}$$

$$\therefore\ \text{one strand has }\frac{1200000}{19}=63158\ \mathrm{CM}$$

$$d_{\mathrm{mil}}=\sqrt{\mathrm{CM}}=\sqrt{63158}=250\,\text{mil}=0.25\,\text{in.}$$

$$d_{\mathrm{mil}}=\sqrt{\mathrm{CM}}=\sqrt{10,382}=102\,\text{mil}=0.102\,\text{in.}$$

$$37\text{ strands have }300\,\mathrm{MCM}=300000\,\mathrm{CM}$$

$$\therefore\ \text{one strand has }\frac{300000}{37}=8108\ \mathrm{CM}$$

$$d_{\mathrm{mil}}=\sqrt{\mathrm{CM}}=\sqrt{8108}=90\,\text{mils}=0.09\,\text{in.}$$

$$R=\frac{KL}{\mathrm{CM}}\quad\Rightarrow\quad L=\frac{R\times\mathrm{CM}}{K}$$

Copper has a K value of 10.4

$$L=\frac{(4000)(2.5)}{10.4}=962\,\text{ft}$$

$$0.1\,\text{in.}=100\,\text{mil}$$

$$A_{\mathrm{CM}}=d^2=(100)^2=10000\ \mathrm{CM}$$

Aluminum has a K value of 17.

$$R=\frac{KL}{\mathrm{CM}}=\frac{17(100)}{10000}=0.17\,\Omega$$

First, calculate the resistance of the wire:

$$R=\frac{KL}{\mathrm{CM}}=\frac{(260)(25)}{404}=16\,\Omega$$

Next, calculate the current:

$$I=\frac{E}{R}=\frac{120}{16}=7.5\,\text{A}$$