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Chapter 30: Problem 52

The transmission of electric power occurs at the highest possible voltage to reduce losses. By how much could the power loss be reduced by raising the voltage by a factor of \(10 ?\)

Short Answer

Expert verified
Answer: The power loss could be reduced by 1000 times.

Step by step solution

01

1. Understanding the relation between power loss and transmission voltage

The power loss in transmission lines is given by the formula: \(P_{loss}=\frac{I^2R}{V}\), where \(I\) is the current, \(R\) is the resistance, and \(V\) is the voltage. From this formula, we can clearly see that power loss is inversely proportional to the voltage. Which means, if we increase the voltage, the power loss will decrease.
02

2. Calculate the new power loss with increased voltage

Let's say, the initial power loss is \(P_{loss1}\) and the new power loss after raising the voltage is \(P_{loss2}\). The voltage is raised by a factor of 10, so the new voltage is \(V_{new} = 10V\) where \(V\) is the initial voltage. Now, we need to find the relation between \(P_{loss1}\) and \(P_{loss2}\). We know the formula for power loss is \(P_{loss}=\frac{I^2R}{V}\), so we can write the power loss equations in terms of initial voltage \(V\) and new voltage \(V_{new}\) as follows: \(P_{loss1}=\frac{I_{1}^2R}{V}\) \(P_{loss2}=\frac{I_{2}^2R}{V_{new}}\)
03

3. Calculate the power loss reduction factor

As we have the relations between the initial and new power losses, let's find how much the power loss is reduced by. We will divide the initial power loss by the new power loss to find the reduction factor: \(reduction\_factor = \frac{P_{loss1}}{P_{loss2}}\) From the power loss equations we have: \(reduction\_factor = \frac{\frac{I_{1}^2R}{V}}{\frac{I_{2}^2R}{10V}}\) Note that the resistance (\(R\)) and the power (\(V\)) are constant, so let's rewrite the reduction factor in terms of the current: \(reduction\_factor=\frac{I_{1}^2}{I_{2}^2}*\frac{10}{1}\) To simplify, assume the initial and new power are the same, then: \(P_{init}=IV=10IV_{new}\) From this equation we have: \(I_{new}= \frac{1}{10}I\)
04

4. Find the reduction factor

Now, we can plug the relation between \(I_{1}\) and \(I_{2}\) into the reduction factor equation: \(reduction\_factor=\frac{I_{1}^2}{(\frac{1}{10}I_{1})^2}*\frac{10}{1}\) Simplify the relation: \(reduction\_factor=\frac{I_{1}^2}{\frac{1}{100}I_{1}^2}*10\) \(reduction\_factor=100*10\)
05

5. Final answer

The power loss could be reduced by a factor of: \(reduction\_factor=1000\). By raising the voltage by a factor of 10, the power loss could be reduced by 1000 times.

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