Question

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

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

Step-by-step solution

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.

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: \(P_{loss1}=\frac{I_{1}^2R}{V}\) \(P_{loss2}=\frac{I_{2}^2R}{V_{new}}\)

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: \(I_{new}= \frac{1}{10}I\)

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: \(reduction\_factor=\frac{I_{1}^2}{\frac{1}{100}I_{1}^2}*10\) \(reduction\_factor=100*10\)

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.

Key concepts

Power Loss Reduction

To understand how power loss reduction works in electric power transmission, it's important to recognize the relationship between power loss, current, and voltage. Power loss in electrical transmission lines is calculated using the formula:\[P_{loss} = \frac{I^2R}{V}\]where:

• \(P_{loss}\) is the power loss
• \(I\) is the current
• \(R\) is the resistance
• \(V\) is the voltage

Generally, lower power loss is desirable as it indicates a more efficient transmission of electricity. From the formula, we see that power loss is inversely proportional to voltage, meaning as the voltage increases, the power loss decreases. This concept explains why high voltage is preferred for the efficient transmission of electricity.
If we increase the transmission voltage by a factor of 10, the power loss is reduced as much as 1000 times due to the squared relationship between current and power in the equation. Therefore, boosting the voltage has a substantial impact in minimizing power loss and improving the efficiency of power transmission systems.

Transmission Voltage Increase

Transmission voltage is a key factor in determining the efficiency of delivering electrical power over long distances. High voltage helps to decrease the amount of current needed to transfer a specific amount of power. This concept can be better understood by revisiting the basic formula for electric power:\[P = IV\]where \(P\) is power, \(I\) is current, and \(V\) is voltage.
By rearranging this equation, we see that for a fixed power output, an increase in voltage results in a reduction in current. Let's consider when voltage is increased by a factor of 10. In this scenario, the current required diminishes significantly. This reduction translates into much lower power losses since losses are proportional to the square of the current \((I^2)\).

• Higher voltage means lower current
• Lower current results in reduced power loss
• Higher efficiency in power transmission systems

This is why electrical systems prioritize transmitting at higher voltages whenever feasible, as it maximizes efficiency and reduces substantial energy losses over the transmission network.

Resistance in Transmission Lines

Resistance within transmission lines is an integral factor when discussing power transmission. Every transmission line has a specific resistance \(R\), which is a measure of how much it opposes the flow of electric current. A higher resistance implies more power loss, as expressed in the formula:\[P_{loss} = \frac{I^2R}{V}\]Understanding how resistance and line losses work is critical for efficient power delivery. Here are some key points:

• Lower resistance results in less power loss
• Resistance is usually inherent to the material and length of the transmission line
• Engineers aim to optimize materials and designs to minimize resistance

When designing transmission systems, minimizing resistance is just as important as managing voltage and current. Cable materials and their cross-sectional areas are carefully selected to adequately support current while keeping resistance low. This balance ensures efficient energy transfer, reducing waste in terms of power lost as heat due to resistance.