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A generating station is producing \(1.2 \times 10^{6} \mathrm{~W}\) of power that is to be sent to a small town located \(7.0 \mathrm{~km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of length of \(5.0 \times 10^{-2} \Omega / \mathrm{km} .\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{~V} .\) (b) A \(100: 1\) step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?

Short Answer

Expert verified
(a) Without a transformer, 700,000 W is used to heat the wires. (b) With a transformer, 70 W is used to heat the wires.

Step by step solution

01

Calculate Wire Resistance

First, calculate the total resistance of the transmission wires. The resistance per kilometer is given as \( 5.0 \times 10^{-2} \Omega/\text{km} \) and the distance is 7.0 km each way, for 14 km total. Thus, the total resistance is \( R = 14 \times 5.0 \times 10^{-2} = 0.7 \Omega \).
02

Calculate Current without Transformer

Use the formula for electrical power \( P = VI \), where \( P = 1.2 \times 10^6 \text{ W} \) and \( V = 1200 \text{ V} \). Solve for the current \( I \) using \( I = \frac{P}{V} = \frac{1.2 \times 10^6}{1200} = 1000 \text{ A} \).
03

Calculate Power Loss without Transformer

The power loss due to the resistance in the wires is given by \( P_\text{loss} = I^2R \). Substituting the values, we get \( P_\text{loss} = (1000)^2 \times 0.7 = 700,000 \text{ W} \).
04

New Voltage with Transformer

With a 100:1 step-up transformer, the new transmission voltage is \( 1200 \times 100 = 120,000 \text{ V} \).
05

Calculate Current with Transformer

Recalculate the current with the step-up voltage using \( I = \frac{P}{V} = \frac{1.2 \times 10^6}{120,000} = 10 \text{ A} \).
06

Calculate Power Loss with Transformer

Calculate the power loss in the wires with the new current: \( P_\text{loss} = I^2R = (10)^2 \times 0.7 = 70 \text{ W} \).

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Key Concepts

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

Transformer Efficiency
Transformers are essential devices in electrical power systems for modifying voltage levels effectively. A key performance indicator for transformers is their efficiency. Efficiency in transformers is the ratio of output power to input power, expressed usually as a percentage. For a transformer with high efficiency, almost all the input power is transferred to the output, with minimal loss.
Transformer efficiency is affected by several factors, such as core losses and copper losses. Core losses occur in the transformer's iron core due to hysteresis and eddy currents. Copper losses arise from the resistance in the windings of the transformer. In general, maintaining high transformer efficiency is crucial for reducing energy loss in power systems.
Electrical Resistance
Electrical resistance is a measure of the opposition that a material offers to the flow of electric current, usually measured in ohms (Ω). Resistance depends on the material's properties, its temperature, and its dimensions. For instance in wires used for power transmission, the resistance might increase with the length of the wire or decrease with a larger cross-sectional area.
  • Resistance per unit length is crucial in calculating total resistance over long distances, like the wire resistance per kilometer given in the exercise.
  • This principle allows for the estimation of power loss through the transmission wires, highlighting the importance of minimizing resistance in economic and efficient power networks.
Power Loss Calculation
Calculating power loss is essential for understanding the efficiency of power transmission. Power loss in a transmission line can be determined using the formula \( P_{\text{loss}} = I^2R \), where \( I \) is the current through the wires, and \( R \) is the resistance of the wires.
High currents lead to significant power losses, as shown by the exercise where power loss is 700,000 W without a transformer, and only 70 W after using a step-up transformer. This reduction occurs because raising the voltage substantially decreases the current for the same power, thereby reducing losses.
Step-up Transformer
A step-up transformer increases voltage from the primary side (input) to the secondary side (output). It is commonly used in power distribution to boost voltage levels and thus reduce current levels, minimizing energy losses across long transmission lines.
In the exercise, the 100:1 step-up transformer increases the transmission voltage from 1200 V to 120,000 V. This increase drastically reduces the current needed to transmit the same amount of power, thereby minimizing power losses significantly in the transmission wires.
Ohm's Law
Ohm's Law is a fundamental principle in electricity that shows a relationship between voltage, current, and resistance. It is expressed as \( V = IR \), where \( V \) is the voltage across the device, \( I \) is the current through the device, and \( R \) is the resistance.
  • Ohm's Law helps calculate how much current will flow through a circuit given the voltage and resistance, which is crucial during power transmission design decisions.
  • Understanding this law is vital for calculating power distribution as shown in this exercise, where using a higher voltage (and thereby lower current) leads to decreased power loss.

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Most popular questions from this chapter

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