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Chapter 22: Problem 82

Concept Questions In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a wall receptacle. (a) How is the power delivered by the receptacle to the primary related to the power delivered by the secondary to the picture tube? Give your answer in the form of an equation, and explain what assumptions are implied when this equation is used. (b) How is the turns ratio of the transformer related to the currents in the primary and the secondary? (c) How is the turns ratio of the transformer related to the voltage across the primary and the voltage across the secondary? (d) Express the turns ratio \(N_{\mathrm{s}} /\) \(N_{\mathrm{p}}\) of the transformer in terms of the power \(P\) used by the picture tube, the voltage \(V_{\mathrm{p}}\) across the primary, and the current \(I_{\mathrm{s}}\) in the secondary. Problem The primary of the transformer is connected to a \(120-\mathrm{V}\) receptacle. The picture

Short Answer

Expert verified
The power delivered by the receptacle is equal to the power used by the picture tube assuming 100% efficiency: \( P_p = P_s \). The turns ratio \( \frac{N_s}{N_p} \) relates to current as \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \) and voltage as \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), and can be expressed as \( \frac{N_s}{N_p} = \frac{P}{V_p \cdot I_s} \).

Step by step solution

01

Understanding Power Equation

In an ideal transformer, the power in the primary coil (input) is equal to the power in the secondary coil (output), assuming 100% efficiency. This relationship can be expressed as \( P_p = P_s \), where \( P_p \) is the power in the primary and \( P_s \) is the power in the secondary.
02

Relating Turns Ratio to Current

The turns ratio of the transformer, which is the ratio of the number of turns in the secondary coil to the number of turns in the primary coil, is related to the currents. This relationship is \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \), where \( I_s \) is the secondary current, \( I_p \) is the primary current, and \( N_s \), \( N_p \) are the number of turns in the secondary and primary, respectively.
03

Relating Turns Ratio to Voltage

The turns ratio of the transformer is also related to the voltages across its coils. This relationship is given by the equation \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \), where \( V_s \) is the secondary voltage and \( V_p \) is the primary voltage.
04

Expressing Turns Ratio in Terms of Power and Current

To express the turns ratio in terms of the power used by the picture tube \( P \), the primary voltage \( V_p \), and the secondary current \( I_s \), we start with the power formula \( P_s = V_s \cdot I_s = P \). Using the relationship \( V_s = \frac{N_s}{N_p} V_p \) and \( P_p = P_s \), we can express the turns ratio: \( \frac{N_s}{N_p} = \frac{P}{V_p \cdot I_s} \).

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Key Concepts

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

Turns Ratio
The turns ratio in a transformer is crucial for understanding the relationship between primary and secondary coils. In simple terms, the turns ratio is the ratio of the number of turns in the secondary coil (\(N_s\)) to the number of turns in the primary coil (\(N_p\)). This ratio is expressed as \(\frac{N_s}{N_p}\).

The turns ratio directly affects both the current and voltage in the transformer. This means that by changing the number of turns in each coil, we can control how voltage is increased or decreased when moving from the primary side to the secondary side.

A high turns ratio implies a high increase in voltage and vice versa. Understanding this ratio is essential for designing transformers to match specific power needs in circuits.
Power Equation
In transformers, the power equation is a decisive principle. For an ideal transformer, the input power in the primary coil is equal to the output power in the secondary coil. This relationship is expressed by the equation \( P_p = P_s \), where \( P_p \) is the power in the primary and \( P_s \) is the power in the secondary.

This equation assumes 100% efficiency, meaning all power is perfectly transferred between coils without any losses. Real-world transformers are not perfectly efficient due to energy losses like heat, but for simplified calculations, assuming an ideal transformer helps in understanding foundational concepts.

From this equation and by knowing either the current or voltage in any one coil, it is possible to calculate the corresponding value in the other coil by rearranging the formula.
Current and Voltage Relationship
A transformer's operation hinges on the careful balance between current and voltage between its coils. The relationship between the turns ratio and current, \( \frac{I_s}{I_p} = \frac{N_p}{N_s} \), indicates that as the number of turns on the secondary coil increases, the current decreases for a fixed amount of power.

Conversely, the voltage relationship is defined by \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \). This equation states that an increase in the number of turns in the secondary coil scales up the voltage output.

These formulas show how, in transforming electrical energy, an increase in voltage results in a decrease in current, and an increase in current results in a decrease in voltage, adhering to the constant power rule for ideal transformers.
Ideal Transformer
The concept of an ideal transformer is a theoretical model used in physics to simplify the understanding of how transformers work. An ideal transformer is assumed to operate with perfect efficiency (100%), meaning no energy is lost during the transfer process between the primary and secondary coils.

In practice, this means for any given amount of power input into the primary coil, an equal amount of power is transferred to the secondary coil.

By using the ideal transformer model, we can easily derive relationships dealt with in transformers, such as the constant power relation \( P_p = P_s \), and use these to explore further more practical scenarios.

While real transformers include losses due to heat dissipation and other factors, the ideal transformer provides a crucial baseline for understanding electric transformers' operations.

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Most popular questions from this chapter

Concept Questions A constant current \(I\) exists in a solenoid whose inductance is \(L .\) The current is then reduced to zero in a certain amount of time. (a) If the wire from which the solenoid is made has no resistance, is there a voltage across the solenoid during the time when the current is constant? (b) If the wire from which the solenoid is made has no resistance, is there an emf across the solenoid during the time that the current is being reduced to zero? (c) Does the solenoid store electrical energy when the current is constant? If so, express this energy in terms of the current and the inductance. (d) When the current is reduced from its constant value to zero, what is the rate at which energy is removed from the solenoid? Express your answer in terms of the initial current, the inductance, and the time during which the current goes to zero. Problem A solenoid has an inductance of \(L=3.1 \mathrm{H}\) and carries a current of \(I=15 \mathrm{~A}\). (a) If the current goes from 15 to \(0 \mathrm{~A}\) in a time of \(75 \mathrm{~ms}\), what is the emf induced in the solenoid? (b) How much electrical energy is stored in the solenoid? (c) At what rate must the electrical energy be removed from the solenoid when the current is reduced to zero in \(75 \mathrm{~ms} ?\)

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?

A circular coil \((950\) turns, radius \(=0.060 \mathrm{~m})\) is rotating in a uniform magnetic field. At \(t=0 \mathrm{~s}\), the normal to the coil is per pendicular to the magnetic field. At \(t=0.010 \mathrm{~s}\) the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made oneeighth of a revolution. An average emf of magnitude \(0.065 \mathrm{~V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a transformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. (a) Is the transformer a step-up or a step-down transformer? (b) Is the current that goes through the batteries greater than, equal to, or smaller than the current coming from the wall socket? (c) If the transformer has a negligible resistance, is the electric power delivered to the batteries greater than, equal to, or less than the power coming from the wall socket? In all cases, provide a reason for your answer. the batteries of a laptop computer are rated at \(9.0 \mathrm{~V}\), and a current of \(225 \mathrm{~mA}\) is used to charge them. The wall socket provides a voltage of \(120 \mathrm{~V}\). (a) Determine the turns ratio of the transformer, (b) What is the current coming from the wall socket? (c) Find the power delivered by the wall socket and the power sent to the batteries. Be sure your answers are consistent with your answers to the Concept Questions.

Interactive Solution \(\underline{22.39}\) at provides one model for solving this problem. The maximum strength of the earth's magnetic field is about \(6.9 \times 10^{-5} \mathrm{~T}\) near the south magnetic pole. In principle, this field could be used with a rotating coil to generate 60.0 Hz ac electricity. What is the minimum number of turns (area per turn \(=0.022 \mathrm{~m}^{2}\) ) that the coil must have to produce an rms voltage of \(120 \mathrm{~V} ?\)
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