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Chapter 22: Problem 69

A generator is connected across the primary coil ( \(N_{p}\), turns) of a transformer, while a resistance \(R_{2}\) is connected across the secondary coil \(\left(N_{\mathrm{s}}\right.\) turns). This circuit is equivalent to a circuit in which a single resistance \(R_{1}\) is connected directly across the generator, without the transformer. Show that \(R_{1}=\left(N_{\mathrm{p}} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with \(\mathrm{Ohm}^{\prime}\) s law as applied to the secondary coil.

Short Answer

Expert verified
The equivalent resistance is \( R_1 = \left(\frac{N_p}{N_s}\right)^2 R_2 \).

Step by step solution

01

Understanding Ohm's Law in the Secondary Coil

In the secondary coil, Ohm's law can be applied as follows: the voltage across the secondary coil \( V_s \) is equal to the current \( I_s \) flowing through \( R_2 \) times the resistance, i.e., \( V_s = I_s R_2 \). This is the initial expression we'll use to connect the resistance seen by the generator to the load resistance \( R_2 \).
02

Relate Voltages and Turns in the Transformer

In a transformer, the voltages across the coils are related to the turns in each coil. The relationship is given by \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \). This formula relates the primary voltage \( V_p \) to the secondary voltage \( V_s \) through the turns ratio \( \frac{N_p}{N_s} \).
03

Express the Secondary Voltage in Terms of the Primary Voltage

Using the relation from Step 2, express \( V_s \) in terms of \( V_p \): \( V_s = V_p \frac{N_s}{N_p} \). This expression helps us to substitute \( V_s \) in terms of quantities involving the primary coil.
04

Substitute into Ohm's Law

Substitute the expression for \( V_s \) from Step 3 into \( V_s = I_s R_2 \). Replace \( V_s \) with \( V_p \frac{N_s}{N_p} \), giving us \( V_p \frac{N_s}{N_p} = I_s R_2 \).
05

Find the Equivalent Resistance Relation

To find equivalent resistance \( R_1 \) seen from the primary side, consider \( I_p = \frac{V_p}{R_1} \). The power input to the primary coil (\( V_p I_p \)) is equal to power output from the secondary coil (\( V_s I_s \)). Substituting \( V_s = V_p \frac{N_s}{N_p} \) and \( I_s = \frac{V_p}{R_1} \frac{N_s}{N_p} \):\( V_p \left(\frac{V_p}{R_1}\right) = \left(V_p \frac{N_s}{N_p}\right) \left(\frac{V_p}{R_1 R_2}\right) \frac{N_s}{N_p} \).
06

Simplify to Find \( R_1 \) in Terms of \( R_2 \)

Simplify the equation in Step 5 to solve for \( R_1 \):\( \frac{V_p^2}{R_1} = \frac{N_s^2}{N_p^2} \frac{V_p^2}{R_2} \).Cancel out \( V_p^2 \) and rearrange to find \( R_1 \):\( R_1 = \left(\frac{N_p}{N_s}\right)^2 R_2 \).
07

Confirm the Derivation

Our derived formula, \( R_1 = \left(\frac{N_p}{N_s}\right)^2 R_2 \), shows that the equivalent resistance \( R_1 \) seen from the primary side is scaled by the square of the turns ratio \( \left(\frac{N_p}{N_s}\right)^2 \) with respect to \( R_2 \). Thus, the relationship has been successfully proven using Ohm's Law and transformer principles.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental concept in electrical circuits that expresses the relationship between voltage, current, and resistance. It states that the voltage across a resistor is equal to the product of the resistance and the current flowing through it. In mathematical terms, this is expressed as: \( V = I \times R \) where \( V \) represents the voltage, \( I \) is the current, and \( R \) is the resistance. In the context of a transformer, Ohm's Law is applied to the secondary coil to find the voltage across it when a resistance \( R_2 \) is connected. This gives us the equation \( V_s = I_s \times R_2 \).
  • This equation helps to determine how much voltage is present across the load resistance based on the current flowing through it.
  • The understanding of Ohm's Law in this scenario is crucial to begin deducing the equivalent resistance on the primary side of the transformer.
voltage turns ratio
The voltage turns ratio is a key principle in transformer operation that relates the number of turns in the primary and secondary coils to the respective voltages across these coils. The formula given by this principle is: \[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] where \( V_p \) and \( V_s \) are the voltages across the primary and secondary coils, and \( N_p \) and \( N_s \) are the number of turns in the primary and secondary coils, respectively. This ratio indicates how a transformer steps up or steps down voltage:
  • If \( N_p > N_s \), the transformer steps down the voltage, providing a lower voltage in the secondary coil compared to the primary.
  • Conversely, if \( N_p < N_s \), the transformer steps up the voltage, providing a higher voltage in the secondary coil.
  • This relationship is crucial for calculating the voltage changes in the transformer and further understanding the equivalent resistance calculation.
transformer principles
Transformers operate on the principles of electromagnetic induction to transfer electrical energy from one circuit to another. A primary coil receives a varying input voltage which induces an electromotive force (EMF) in the secondary coil. The key transformer principles are based on:
  • Electromagnetic Induction: A changing magnetic field within the transformer's core due to alternating current in the primary coil results in a voltage being induced in the secondary coil.
  • Conservation of Energy: The input power (product of voltage and current) in the primary must equal the output power in the secondary, minus any losses. This principle is reflected in the equation: \[ V_p \times I_p = V_s \times I_s \]
  • Transformers also change the effective impedance of circuits. The impedance reflected from the secondary to the primary is affected by the square of the turns ratio: \[ R_1 = \left(\frac{N_p}{N_s}\right)^2 R_2 \] This allows simplifying complex circuits into more manageable equivalents.
These fundamental principles allow transformers to efficiently step voltages up or down and adapt impedances for various applications.

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Most popular questions from this chapter

During a 72-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 6.0-mA current in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12\(\Omega .\) The mutual inductance between the two coils is 3.2 \(\mathrm{mH}\) . What is the change in the primary current?

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the trans- former is connected to a \(120-\mathrm{V}\) receptacle on a wall. The picture tube of the television set uses 91 \(\mathrm{W}\) , and there is 5.5 \(\mathrm{mA}\) of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio \(N_{s} / N_{\mathrm{p}}\) of the transformer.

The magnetic flux that passes through one turn of a 12 -turn coil of wire changes to 4.0 \(\mathrm{from} 9.0 \mathrm{Wb}\) in a time of 0.050 \(\mathrm{s}\) . The average induced current in the coil is 230 \(\mathrm{A}\) . What is the resistance of the wire?

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of 0.46 V to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a handgun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

The coil of a generator has a radius of 0.14 \(\mathrm{m}\) . When this coil is unwound, the wire from which it is made has a length of 5.7 \(\mathrm{m}\) . The magnetic field of the generator is \(0.20 \mathrm{T},\) and the coil rotates at an angular speed of 25 \(\mathrm{rad} / \mathrm{s}\) . What is the peak emf of this generator?
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