Electricity Help: Get Answers to Your Questions

[BLT]

Hi first post here. I will state the question then point out my thoughts.

Question
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Energy is delivered to your house via high voltage power lines which are transformed to low voltage in your back yard. Imagine that the energy used to give you energy was created at the Braidwod Nuclear Power Plant located 50 miles south of here. The energy is delivered over 120,000 Volt lines. The lines that are used typically have a cross sectional area of (3 x 10^ -4 meter^2). They are usually made of coper which has a resistivity of (1.7 x 10^ -8 ohm-meter). The average American Household uses about 1000 kilowatt-hours of energy in a month. Determine the following:

a) What must be the average current which travels from Braidwood to your backyard?

b) What would be the total resistance of the wire from Braidwood to your backyard?

c) How much energy is lost in one month by the heating of the wire in the process of bringing energy to your house?

Extra Credit: How much energy would be lost in a month if the energy was delivered at a potential of 120 volts?

My Thoughts
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My approach for a) was to use the I = sqr(P/R). I used R=(pL)/A formula to solve for R. My concern is whether I correctly used R...

Which leads to my problem for question b). 120,000 Volt lines are used. If R=(pL)/A for one Volt line, then how would I find the total resistance with 120,000 Volt lines used? Do I treat them as in series or in parallel?

As for C) I have absolutely no clue. I would guess that I subtract electrical energy from total energy? But I can't even distinguish the two equationally.

I think I can solve the Extra Credit once I know how to do C). But any help is of course welcome.
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This is a long series of questions, but I would appreciate any help. Thanks in advance.

 

[Chen]

The total resistance of resistors in parallel is:
$$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$$
If the resistors are all requal then:
$$\frac{1}{R_T} = \frac{1}{R_i} + \frac{1}{R_i} + ... = \frac{N}{R_i}$$
Therefore the total resistance is:
$$R_T = \frac{R_i}{N}$$

As for (c), the thermal energy that is lost in wires is the power times time:
$$\Delta E_T = P\Delta t$$
Note that P, the power, is of the wires and not of the house. You have the resistance of the wires from (b) and the current from (a). As for the time, it is 30 (or 31?) days times 24 hours times 60 minutes times 60 seconds.

 

[M98Ranger]

Hi there, welcome to the forum! Let's take a look at your questions one by one and try to find some answers.

a) To find the average current, we can use the formula I = P/V, where P is the power (in watts) and V is the voltage (in volts). Since we know that the average American household uses 1000 kilowatt-hours (kWh) of energy in a month, we can convert this to watts by multiplying by 1000 (since 1 kWh = 1000 watts). This gives us 1,000,000 watts. Now, we need to convert this to joules per second (since watts are a measure of energy per second). To do this, we divide by the number of seconds in a month (30 days x 24 hours x 60 minutes x 60 seconds = 2,592,000 seconds). This gives us an average power of 1,000,000 watts / 2,592,000 seconds = 0.385 joules per second. Now, we can plug this into our formula I = P/V, where P = 0.385 joules per second and V = 120,000 volts. This gives us an average current of 0.385/120,000 = 3.2 x 10^-6 amps.

b) To find the total resistance, we need to use the formula R = pL/A, where p is the resistivity of the material (in ohm-meters), L is the length of the wire (in meters), and A is the cross-sectional area of the wire (in meters squared). In this case, we know that the resistivity of copper is 1.7 x 10^-8 ohm-meters, and the cross-sectional area is 3 x 10^-4 meters squared. However, we are not given the length of the wire, so we cannot find the total resistance. We can only find the resistance per meter of wire. If we assume that the length of the wire is 50 miles (80.5 km), we can convert this to meters (1 mile = 1609 meters) and get a total resistance of 1.3 x 10^-6 ohms.

c) To find the energy lost in one month due to the heating of the wire, we need to use the formula E = I^2Rt,