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Chapter 19: Problem 78

A power plant transmits \(150 \mathrm{~kW}\) of power to a nearby town through wires that have a total resistance of \(0.25 \Omega\). What percentage of the power is dissipated as heat in the wire if the power is transmitted at (a) \(220 \mathrm{~V}\) and (b) \(22 \mathrm{kV} ?\)

Short Answer

Expert verified
At 220 V, 77.35% of the power is lost as heat. At 22 kV, only 0.0078% is lost.

Step by step solution

01

Power Loss Formula

To find the percentage of power lost as heat, we first identify the formula for power loss due to resistance in the wire: \[ P_{loss} = I^2 R \]where \( I \) is the current through the wire and \( R \) is the resistance (0.25 惟).
02

Calculate Current for 220 V

Use the formula for power to find the current at 220 V:\[ P = VI \rightarrow I = \frac{P}{V} \]Substitute \( P = 150 \, \mathrm{kW} = 150,000 \, \mathrm{W} \) and \( V = 220 \, \mathrm{V} \):\[ I = \frac{150,000}{220} = 681.82 \, \mathrm{A} \]
03

Power Loss Calculation at 220 V

Using the current calculated:\[ P_{loss} = (681.82)^2 \times 0.25 \]\[ P_{loss} = 116,025.25 \, \mathrm{W} = 116.025 \, \mathrm{kW} \]
04

Percentage Loss at 220 V

The percentage of power loss is given by:\[ \text{Percentage Loss} = \left( \frac{P_{loss}}{P} \right) \times 100 \]Substitute the known values:\[ \text{Percentage Loss} = \left( \frac{116.025}{150} \right) \times 100 = 77.35\% \]
05

Calculate Current for 22 kV

Now, repeat the process for 22 kV:\[ V = 22 \, \mathrm{kV} = 22,000 \, \mathrm{V} \]\[ I = \frac{150,000}{22,000} = 6.82 \, \mathrm{A} \]
06

Power Loss Calculation at 22 kV

Using the new current calculated:\[ P_{loss} = (6.82)^2 \times 0.25 \]\[ P_{loss} = 11.6555 \, \mathrm{W} \approx 11.66 \, \mathrm{W} \]
07

Percentage Loss at 22 kV

Calculate the percentage of power lost:\[ \text{Percentage Loss} = \left( \frac{11.66}{150,000} \right) \times 100 \]\[ \text{Percentage Loss} \approx 0.0078\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Loss Calculation
Power loss during transmission is a crucial factor to consider in any power distribution system. When electricity travels through wires, some of it is lost as heat due to the resistance of the wire. This lost power is often calculated using the formula for power loss:
  • \( P_{loss} = I^2 R \)
\( I \) is the electric current flowing through the wire, and \( R \) is the resistance of the wire. It's important to minimize power loss to improve transmission efficiency. In our example, we see different power loss results at different voltages due to changes in current.
Electric Current
Electric current, denoted as \( I \), is the flow of electric charge. In a power transmission system, calculating the current is key to understanding power flow. The current is directly related to the voltage and power transmitted via the equation:
  • \( P = VI \)
  • Rearrange to find \( I = \frac{P}{V} \)
This shows that current inversely relates to voltage if power remains constant. In the exercise, lowering the voltage (from 22 kV to 220 V) increased the current significantly, impacting power loss.
Resistance
Resistance, symbolized by \( R \), is a property that opposes the flow of electric current in a conductor. The resistance of transmission wires is a critical component determining power loss:
  • Calculated in ohms (惟)
  • Higher resistance means more power loss as heat
In the exercise, with a resistance of \(0.25 \Omega\), varying the current had a significant effect on the power lost as heat, showcasing why resistance management is key in power network design.
Ohm's Law
Ohm's Law forms the foundation for understanding electrical circuits. It relates voltage \( V \), current \( I \), and resistance \( R \) by the equation:
  • \( V = IR \)
This relationship is crucial in calculating how changes in one affect the others. For example, in the exercise, adjusting the voltage changes the current for the same power. A better grasp of Ohm鈥檚 Law can help explain why high voltages reduce power loss in long-distance transmission.

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Most popular questions from this chapter

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to \(500 \mathrm{~V}\) and produce currents of \(80 \mathrm{~mA}\) (or even larger). A typical pulse lasts for \(10 \mathrm{~ms}\). What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?

A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at \(25 \mathrm{~V}\) for a very short time, usually about \(3.0 \mathrm{~ms}\). (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

A \(4600 \Omega\) resistor is connected across a charged \(0.800 \mathrm{nF}\) capacitor. The initial current through the resistor, just after the connection is made, is measured to be \(0.250 \mathrm{~A}\). (a) What magnitude of charge was initially on each plate of this capacitor? (b) How long after the connection is made will it take before the charge is reduced to \(1 / e\) of its maximum value?

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic-that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a \(5 \mathrm{~mm}\) length of thread between two electrical contacts. The researchers stretched the thread in \(1 \mathrm{~mm}\) increments to more than twice its original length, and then allowed it to return to its original length, again in \(1 \mathrm{~mm}\) increments. Some of the resistance measurements are given in the table: $$\begin{array}{l|cccccccc}\begin{array}{l}\text { Resistance of } \\\\\text { thread }\left(10^{9} \Omega\right)\end{array} & 9 & 19 & 41 & 63 & 102 & 76 & 50 & 24 \\ \hline \begin{array}{l}\text { Length of } \\\\\text { thread }(\mathrm{mm})\end{array} & 5 & 7 & 9 & 11 & 13 & 9 & 7 & 5\end{array}$$ If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area \(A\) of the coating compare when the thread is \(13 \mathrm{~mm}\) long versus the starting length of \(5 \mathrm{~mm} ?\) Assume that the resistivity of the coating remains constant and the coating is uniform along the thread. \(A_{13 \mathrm{~mm}}\) is about A. \(1 / 10 A_{5 \mathrm{~mm}}\) B. \(1 / 4 A_{5 \mathrm{~mm}}\) C. \(2.5 \mathrm{~A}_{5 \mathrm{~mm}}\) D. the same as \(A_{5 \mathrm{~mm}}\)
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