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An ac generator supplies \(20 \mathrm{~A}\) at \(440 \mathrm{~V}\) to a \(10000-\mathrm{V}\) power line through a step-up transformer that has 150 turns in its primary coil. (a) If the transformer is \(95 \%\) efficient (see Exercise 32), how many turns are in the secondary coil? (b) What is the current in the power line?

Short Answer

Expert verified
(a) 3410 turns in the secondary coil. (b) 0.836 A on the power line.

Step by step solution

01

Understanding Transformer Efficiency

First, understand that transformer efficiency is given as \( \eta = \frac{P_\text{out}}{P_\text{in}} \times 100\% \). Here, \( P_\text{in} = V_\text{primary} \times I_\text{primary} \) and \( P_\text{out} = V_\text{secondary} \times I_\text{secondary} \). With an efficiency of 95%, set up the equation \( 0.95 = \frac{V_\text{secondary} \times I_\text{secondary}}{V_\text{primary} \times I_\text{primary}} \).
02

Calculate the Output Power of Transformer

Calculate the input power \( P_\text{in} \) using \( V_\text{primary} = 440 \mathrm{~V} \) and \( I_\text{primary} = 20 \mathrm{~A} \):\[ P_\text{in} = 440 \mathrm{~V} \times 20 \mathrm{~A} = 8800 \mathrm{~W}. \]Using the efficiency \( \eta \, = \, 95\% \), we have:\[ P_\text{out} = 0.95 \times 8800 \mathrm{~W} = 8360 \mathrm{~W}. \]
03

Determine Secondary Voltage and Turns Ratio

The secondary voltage \( V_\text{secondary} \) is given as \( 10000 \mathrm{~V} \). Using the ratio of the turns in the coils \( \frac{N_\text{secondary}}{N_\text{primary}} = \frac{V_\text{secondary}}{V_\text{primary}} \), and given \( N_\text{primary} = 150 \), compute \( N_\text{secondary} \):\[ \frac{N_\text{secondary}}{150} = \frac{10000}{440}. \]\[ N_\text{secondary} = \frac{10000}{440} \times 150 = 3409.09. \]Since the number of turns must be an integer, round this to \( N_\text{secondary} = 3410 \text{ turns}. \)
04

Calculate Current in Power Line

Using the output power \( P_\text{out} = V_\text{secondary} \times I_\text{secondary} = 8360 \mathrm{~W} \) and \( V_\text{secondary} = 10000 \mathrm{~V} \), calculate \( I_\text{secondary} \):\[ I_\text{secondary} = \frac{8360}{10000} = 0.836 \mathrm{~A}. \]The current in the power line is thus \( 0.836 \mathrm{~A} \).

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Key Concepts

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

AC Generator
An AC generator is an essential device used in power generation. It converts mechanical energy into electrical energy using the principles of electromagnetic induction. This device essentially works by rotating a coil in a magnetic field to induce an alternating current (AC).
The output from an AC generator can vary in both voltage and current, making it extremely versatile. In our problem, the AC generator provides an initial supply of 20 A at 440 V. This initial voltage must be stepped-up to meet the demands of a high-voltage power line.
Items to consider when understanding AC generators include:
  • The rotating coil creates alternating current, which changes direction periodically.
  • The frequency of the AC generated depends on the speed of coil rotation.
  • AC generators are widely used in central power stations and smaller industrial applications to meet energy needs.
Step-up Transformer
A step-up transformer is an electrical device that increases the voltage from the primary coil to the secondary coil while reducing the current. It does this using the principle of electromagnetic induction, where the number of turns in each coil affects the voltage level.
In this case, the function of the step-up transformer is to increase the voltage from the generator's 440 V to a 10000 V power line. This is achieved through the difference in the number of turns between the primary and secondary coils.
  • Step-up transformers are essential for efficiently transmitting electrical power over long distances.
  • They help reduce the power losses that occur due to resistance in transmission lines by increasing voltage.
  • The efficiency of a transformer impacts how much of the input power is transferred to the output without losses.
Turns Ratio
The turns ratio of a transformer is a critical factor that determines how the voltage will be transformed from the primary coil to the secondary coil. It is given by the equation:

\( \frac{N_{\text{secondary}}}{N_{\text{primary}}} = \frac{V_{\text{secondary}}}{V_{\text{primary}}} \)

Here, \(N_{\text{secondary}}\) and \(N_{\text{primary}}\) represent the number of turns in the secondary and primary coils, respectively, while \(V_{\text{secondary}}\) and \(V_{\text{primary}}\) represent their respective voltages.
For our exercise, the turns ratio allows us to calculate the exact number of turns needed in the secondary coil to step-up the voltage from 440 V to 10000 V. Using the provided values and calculations, the secondary coil requires 3410 turns.
  • The turns ratio is a direct guide to how voltage levels change between the transformer coils.
  • This ratio sheds light on whether a transformer is a step-up or step-down type.
  • Understanding turns ratio helps in designing transformers for specific functional requirements.
Power Line Current
Power line current is the amount of electrical current flowing through a power line. This current is crucial for the transmission of electrical power to distant locations.
In our problem, the power line current is found after determining the output power and voltage. By using the formula: \(I_{\text{secondary}} = \frac{P_{\text{out}}}{V_{\text{secondary}}} \)
it is established that the current in the power line is approximately 0.836 A.
Factors to consider regarding power line current include:
  • Maintaining lower current levels in high-voltage transmissions reduces the heat and energy losses.
  • Power line current impacts the efficiency of power distribution systems.
  • Proper calculation of line current ensures safe and reliable power delivery.

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Most popular questions from this chapter

Orange light has a wavelength of \(600 \mathrm{nm},\) and green light has a wavelength of \(510 \mathrm{nm}\). (a) What is the difference in frequency between the two types of light? (b) If you doubled the wavelength of both, what type of light would thev become?

The flux through a loop of wire changes uniformly from \(+40 \mathrm{~Wb}\) to \(-20 \mathrm{~Wb}\) in \(1.5 \mathrm{~ms}\). (a) What is the significance of the negative number attached to the final flux value? (b) What is the average induced emf in the loop? (c) If you wanted to double the average induced emf by changing only the time, what would the new time interval be? (d) If you wanted to double the average induced emf by changing only the final flux value, what would it be?

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In \(0.20 \mathrm{~s}\), a coil of wire with 50 loops experiences an average induced emf of \(9.0 \mathrm{~V}\) due to a changing magnetic field perpendicular to the plane of the coil. The radius of the coil is \(10 \mathrm{~cm}\), and the initial strength of the magnetic field is 1.5 T. Assuming that the strength of the field decreased with time, (a) what is the final strength of the field? (b) If the field strength had, instead, increased, what would its final value have been? (c) Explain a method whereby you could, in principle, tell whether the field was increasing or decreasing in magnitude.

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