What Must the Output of the Powerhouse Be to Power the Factory?

[CraZyBryaN]

Hello. Hopefully someone here is able to answer this. I'm not sure if it's easy or hard, but it sure is frustrating... Here is the question...word for word, I would make a sad attempt at a picture, but it would probably be wrong. So...


"A powerhouse near a waterfall has a large DC generator that produces electricity for a factory 0.50 mile away. The energy is transmitted over two cables each with a resistance of 0.25 Ohms/Mile. Given that the factory requires 45kW at a voltage of 110V to run its equipment, what must be the output of the powerhouse"

We haven't talked a lot about resistances yet, so I think I'm probably making mistakes there...but yea. Thanks in advance.

 

[Astronuc]

For DC, power P = V * I, so I = P/V.

The total resistance over the line, R, is given by R = $\rho$ L, where $\rho$ is the resistance per unit length and L is the total length of the line.

Now what is the voltage drop and power loss over the line?

 

[ogb p]

Hello there,

The first step in solving this problem is to understand the concept of power and how it relates to voltage, current, and resistance. Power is the rate at which energy is transferred or consumed, and it is measured in watts (W). In an electric circuit, power is equal to the product of voltage (V) and current (I), or P=VI. We can also use Ohm's law (V=IR) to find the relationship between voltage, current, and resistance.

In this case, we know that the factory requires 45kW of power at a voltage of 110V. Using the formula P=VI, we can rearrange it to solve for current (I), which is equal to 45kW/110V = 409.09A.

Next, we need to take into account the resistance of the two cables that the electricity will travel through. We are given that each cable has a resistance of 0.25 Ohms/Mile, and the distance between the powerhouse and the factory is 0.50 mile. Therefore, the total resistance of the circuit is 0.25 Ohms/Mile x 0.50 mile = 0.125 Ohms.

Now, we can use Ohm's law (V=IR) again to solve for the output of the powerhouse. We know the voltage (110V) and the total resistance (0.125 Ohms), so we can rearrange the formula to solve for current (I). This gives us I=V/R = 110V/0.125 Ohms = 880A.

Finally, we can use the formula P=VI to find the output of the powerhouse. We know the current (880A) and the voltage (110V), so we can solve for power (P). This gives us P=VI = 110V x 880A = 96,800W or 96.8kW.

Therefore, the output of the powerhouse must be 96.8kW in order to provide 45kW of power to the factory 0.50 mile away, taking into account the resistance of the cables. I hope this helps clarify the problem for you. Good luck!