According to the U.S. National Electrical Code, copper wire...

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current $I_{\max }$ for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$ \begin{array}{ccc} \text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\ \hline 14 & 0.163 & 18 \\ 12 & 0.205 & 25 \\ 10 & 0.259 & 30 \\ 8 & 0.326 & 40 \\ 6 & 0.412 & 60 \\ 5 & 0.462 & 65 \\ 4 & 0.519 & 85 \end{array} $$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of $4200 \mathrm{~W}$ of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is $120 \mathrm{~V},$ determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length $42.0 \mathrm{~m}$. At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is $\$ 0.11$ per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current $I_{\max }$ for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge.
$$
\begin{array}{ccc}
\text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\
\hline 14 & 0.163 & 18 \\
12 & 0.205 & 25 \\
10 & 0.259 & 30 \\
8 & 0.326 & 40 \\
6 & 0.412 & 60 \\
5 & 0.462 & 65 \\
4 & 0.519 & 85
\end{array}
$$
(a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of $4200 \mathrm{~W}$ of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is $120 \mathrm{~V},$ determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length $42.0 \mathrm{~m}$. At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is $\$ 0.11$ per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

Key Concepts

Ampacity

Ampacity refers to the maximum current a conductor can safely carry without exceeding its temperature rating due to resistive heating. It is determined by factors such as the conductor’s material properties, cross-sectional area, type of insulation, ambient temperature, and installation conditions. Understanding ampacity is crucial in wiring to ensure safety and prevent overheating, which could lead to insulation failure or even fires.

Wire Gauge

Wire gauge is a standardized system used to describe the diameter of a wire. The gauge number inversely relates to the diameter; that is, a smaller gauge number indicates a thicker wire. This concept is important because the thickness of the wire directly affects its resistance and current-carrying capacity, ensuring that wiring is appropriately sized for its intended electrical load.

Ohm’s Law and Power Relationships

Ohm’s Law (V = IR) and the power equation (P = IV or P = I²R) are foundational concepts in electrical engineering used to relate current, voltage, resistance, and power. These relationships enable the calculation of current from a given power and voltage supply, and subsequently the determination of resistive losses in the conductor. They are essential for designing circuits that operate efficiently and safely.

Resistive Heating and Energy Dissipation

Resistive heating, also known as Joule heating, occurs when current flows through a conductor with resistance, causing energy to be dissipated as heat. This phenomenon is quantified by the formula P = I²R, where losses can impact the overall efficiency of the electrical system. Calculating these losses is important for both ensuring safety and evaluating the economic implications of wiring choices over time.

Energy Consumption and Cost Analysis

Energy consumption is evaluated by considering the power dissipated over time, typically using the kilowatt-hour (kWh) metric for practical cost assessments. When comparing different wiring options, analyzing the energy lost as heat and converting these losses into cost (using the local electricity rate) helps in understanding the economic benefits of using conductors with lower resistance. This analysis informs decisions in both residential and commercial building designs aimed at enhancing efficiency and reducing operating costs.

Transcript

0:00 Hi there.

00:01 So for this problem we are given a table that shows values of the maximum current for several common sizes of wire.

00:11 So the wire gauge is a standard used to describe the diameter of wires.

00:18 So the larger the diameter of a wire, the smaller the wire gauge.

00:25 So for part a of this problem, and we are asked us, what considerations determine the maximum current carrying capacity of household wiring.

00:38 So to find a solution to that, we need first to study the power supply to the house.

00:47 We know that that is given, the power is given by the potential difference times the current.

00:57 Now, the rate at which the electrical energy is dissipated in the wire is given by the product between the square of the current times the resistance.

01:15 We know that the resistance and also is given by the resistivity of the material times its length times the creptional area.

01:27 Now for cooper, that is the one material that we are using for this problem, the resistivity is 1 .20, 1 .72 times 10 to the minus 8 oms per meter.

01:48 So to answer part a, so part a of this problem, so the considerations that we need to take into account are the line voltage, line voltage, the current to be drawn, current, and the wire diameter, so wire diameter.

02:25 And these are the most to be considered in the household wiring.

02:33 Now, for part b of this problem, we are asked, and it says a total of four, i'm going to put that value in here, it says a total, we have a power of 4 ,200 watts, and we are said that if the potential difference across the group of appliances, we are also given the potential difference, that is 120 volts, and we need to determine the gauge of the tiniest, permissible wire that can be used.

03:19 So for that, the first thing that we need to do is to use the following equation as said from the beginning, the power supplied is equal to the potential difference times the current.

03:35 So we can obtain the current for this.

03:38 And that it's going to be the power over the potential difference.

03:43 So we have these two values.

03:45 One is 1 ,4 ,200 watts and the difference in potential is 120 volts.

03:56 So this will give us a value of 35 ampers.

04:02 Now, so the 8 -gauge wire is necessary.

04:10 So for this, we need to compare this value to the one we obtained from, the one that we are supplied in the...

04:17 In the table.

04:22 So we can see that this in here is the one that we can use because it has a maximum value for the current of 40 ampers.

04:36 So that's the one that we can use in here.

04:39 So it will be the eight gauge wire, the one that is necessary.

04:48 So because it can carry up to 40 empers.

04:53 Now, for for part c of this problem, we are asked to suppose that the wire is used in this house is of the gauge found in part b and has a total length of 42 meters.

05:14 So at what rate is energy dissipated in the wires? so to note that we are going to use again the equation for power, but in this case it's given by the square of the current times the resistance.

05:34 And we know that the definition of resistance is, we put that the current square, and the definition of resistance is the resistivity times the length of it over the area.

05:47 We know all of these values.

05:51 Since we are using the eight gauge wire, the one that we said before, we can search in the table for the diameter.

06:02 We know that this is diameter in here.

06:06 0 .326.

06:08 So that's the diameter.

06:17 0 .326.

06:18 And this is in centimeters...

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