Solve Electrical Power Homework: 5.3kW

In summary, using the given equations, the power wasted is 5.3kW when the electricity is delivered at 50,000V instead of 12,000V. This discrepancy in the two methods is due to the assumption that all the voltage is dropped across the power line in the first method, leading to a higher power wasted calculation.
  • #1
Jstew
18
0

Homework Statement


A power station delivers 520 kW of power to a factory through wires of total resistance of 3ohms. How much less power is wasted if the electricity is delivered at 50000V instead of 12000V?

Homework Equations



[tex]V=IR[/tex]
[tex]P=IV=I^{2}R=\frac{V^{2}}{R}[/tex]

The Attempt at a Solution



[tex]\frac{(50000V)^{2}}{3ohms}=8.3x10^{8}W[/tex]

[tex]\frac{(12000V)^{2}}{3ohms}=4.8x10^{7}W[/tex]

[tex]8.3x10^{8}W-4.8x10^{7}W=7.82x10^{8}W[/tex]

The correct solution, verified by a classmate and the answer key, is 5.3kW
What have I done incorrectly?
 
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  • #2
Yes the asnwer is 5.3kW

What you have done is partly correct, but it is the last step. You have not used your other given equation. Try and use this first equation and solve for the unknown then calculate the power wasted by each.

Remember you are looking for power "wasted" so think voltage "wasted" or in more electrical terms voltage drop in the lines from the plant to the factory.

I hope this helps you out.
 
  • #3
Ok, I found the current first and then, using the power equation with current and resistance, found the power wasted. This gives the correct answer of 5.3kW. However, conceptually, I'm still unclear as to why the two methods give different answers.
 
  • #4
The method you used assumes that, for example, all 12,000 V are dropped across the power line and none across the factory (load). In fact, most of the voltage is dropped across the factory and only a small fraction across the power line. Now that you know the current, you can verify this.
 

1. What does 5.3kW mean?

5.3kW stands for 5.3 kilowatts, which is a unit of power. It is equal to 5,300 watts or 5,300 joules per second. In the context of electrical power, it refers to the rate at which energy is consumed or produced.

2. How do you calculate electrical power?

Electrical power is calculated by multiplying the voltage (in volts) by the current (in amps). The resulting value is measured in watts (W) or kilowatts (kW). In the case of 5.3kW, it can be calculated by dividing the power rating (5.3kW) by the voltage (in volts) to find the current (in amps).

3. What is the significance of 5.3kW in electrical systems?

5.3kW is a relatively high power rating and is commonly used to describe the power output of large appliances such as air conditioners, electric ovens, or washing machines. It can also refer to the maximum amount of power that a circuit or electrical device can handle without risking damage or overload.

4. How does 5.3kW compare to other units of power?

5.3kW is equivalent to 5,300 watts or 5.3 kilowatts. To put it into perspective, the average US household uses around 1kW of power at any given time. A 60-watt light bulb consumes 0.06 kilowatts of power. In comparison, a 5.3kW power rating is on the higher end of the power scale.

5. Can 5.3kW power a whole house?

It depends on the size and energy efficiency of the house. 5.3kW may be enough to power a small to medium-sized house with moderate electricity usage. However, larger houses with more appliances and higher energy consumption may require a higher power rating. It is always best to consult a professional electrician to determine the appropriate power rating for your specific household needs.

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