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Chapter 32: Problem 35

A transformer has 500 primary turns and 10 secondary turns. (a) If \(\Delta V_{p}\) is \(120 \mathrm{~V}\) (rms), what is \(\Delta V_{s}\) with an open circuit? (b) If the secondary now has a resistive load of \(15 \Omega\), what are the currents in the primary and secondary?

Short Answer

Expert verified
(a) 2.4 V (b) Primary current: 0.0032 A, Secondary current: 0.16 A

Step by step solution

01

Identify the given values

The problem gives you the following values: - Number of primary turns, \( N_p = 500 \) - Number of secondary turns, \( N_s = 10 \) - Primary voltage, \( \Delta V_p = 120 \, \mathrm{V} \) - Resistive load in the secondary, \( R_s = 15 \, \Omega \)
02

Calculate the secondary voltage with an open circuit

Using the transformer equation for voltage, \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \), we can solve for the secondary voltage, \( V_s \): \[ V_s = V_p \cdot \frac{N_s}{N_p} \] Substitute the values to get: \[ V_s = 120 \, \mathrm{V} \cdot \frac{10}{500} = 2.4 \, \mathrm{V} \]
03

Calculate the secondary current with resistive load

Using Ohm's law, \( I_s = \frac{V_s}{R_s} \), substitute the values to find the current in the secondary: \[ I_s = \frac{2.4 \, \mathrm{V}}{15 \, \Omega} = 0.16 \, \mathrm{A} \]
04

Calculate the primary current

Using the transformer equation for current, \( \frac{I_p}{I_s} = \frac{N_s}{N_p} \), we can solve for the primary current, \( I_p \): \[ I_p = I_s \cdot \frac{N_s}{N_p} \] Substitute the values to get: \[ I_p = 0.16 \, \mathrm{A} \cdot \frac{10}{500} = 0.0032 \, \mathrm{A} \]

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Key Concepts

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

Transformer Turns Ratio
The transformer turns ratio is a fundamental concept in understanding how transformers operate. It relates the number of turns in the primary coil (N_p) to the number of turns in the secondary coil (N_s). This ratio determines how voltage and current are transformed between the primary and secondary sides of the transformer. The relationship is given by the equation:\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]Where V_p is the primary voltage and V_s is the secondary voltage. For our problem, with N_p = 500 turns and N_s = 10 turns, the ratio \frac{N_p}{N_s} is 50:1. This means the primary voltage is stepped down by a factor of 50 to give the secondary voltage.
Ohm's Law
Ohm's Law is crucial in both AC and DC circuits. It states that the current through a conductor between two points is directly proportional to the voltage across the two points. The law is mathematically expressed as:\[ I = \frac{V}{R} \]Where I is the current, V is the voltage, and R is the resistance. To calculate the current in the secondary coil of our transformer problem, we use the secondary voltage V_s = 2.4 \text{V} and the resistive load R_s = 15 \text{Ohms}. Plugging in these values, we find the current:\[ I_s = \frac{2.4}{15} = 0.16 \text{A} \]
Primary and Secondary Voltage
Transformers work on the principle of electromagnetic induction to convert voltages from one level to another. When an AC voltage is applied to the primary coil, it creates a magnetic field that induces a voltage in the secondary coil. The ratio of the primary voltage (V_p = 120V) to the secondary voltage (V_s) is determined by the turns ratio. For our example, this ratio is:\[ V_s = V_p \times \frac{N_s}{N_p} \]Using the values N_p = 500 and N_s = 10, the secondary voltage is:\[ V_s = 120 \times \frac{10}{500} = 2.4V \]
Primary and Secondary Current
The current transformation in a transformer also follows the turns ratio, but in an inverse manner compared to voltage. The relationship is given by:\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \]Using the secondary current I_s = 0.16A from our calculations and the turns ratio, we find the primary current I_p:\[ I_p = I_s \times \frac{N_s}{N_p} \]Substituting the numbers:\[ I_p = 0.16 \times \frac{10}{500} = 0.0032A \]This means the current in the primary coil is much lower than in the secondary because the voltage is much higher.
Resistive Load
A resistive load in the context of transformers refers to a resistance connected across the secondary winding that draws current. When we calculate the current in the secondary coil, knowing the resistance value is crucial. From Ohm's Law, the current I_s can be found as:\[ I_s = \frac{V_s}{R_s} \]For our problem, with a secondary voltage V_s = 2.4V and a resistive load R_s = 15\text{Ohms}, the current is:\[ I_s = \frac{2.4}{15} = 0.16A \].This serves as a practical example of how transformers behave when connected to different types of loads, including purely resistive ones.

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Most popular questions from this chapter

At time \(t=0 \mathrm{~ms}\), a \(45.0 \mathrm{~V}\) potential difference is suddenly applied to a coil with \(L=50.0 \mathrm{mH}\) and \(R=\) \(180 \Omega\). At what rate is the current increasing at \(t=1.20 \mathrm{~ms}\) ?

The current in an \(R L\) circuit drops from \(1.0 \mathrm{~A}\) to \(10 \mathrm{~mA}\) in the first second following removal of the battery from the circuit. If \(L\) is \(10 \mathrm{H}\), find the resistance \(R\) in the circuit.

Measurements in mines and boreholes indicate that the Earth's interior temperature increases with depth at the average rate of \(30 \mathrm{C}^{\circ} / \mathrm{km}\). Assuming a surface temperature of \(10^{\circ} \mathrm{C}\), at what depth does iron cease to be ferromagnetic? (The Curie temperature of iron varies very little with pressure.)

An electron with kinetic energy \(K\) travels in a circular path that is perpendicular to a uniform magnetic field, the electron's motion is subject only to the force due to the field. (a) Show that the magnetic dipole moment of the electron due to its orbital motion has magnitude \(\mu=K /|\vec{B}|\) and that it is in the direction opposite that of \(\vec{B}\). (b) What are the magnitude and direction of the magnetic dipole moment of a positive ion with kinetic energy \(K_{\text {ion }}\) under the same circumstances? (c) An ionized gas consists of \(5.3 \times 10^{21}\) electrons \(/ \mathrm{m}^{3}\) and the same number density of ions. Take the average electron kinetic energy to be \(6.2 \times 10^{-20} \mathrm{~J}\) and the average ion kinetic energy to be \(7.6 \times 10^{-21} \mathrm{~J} .\) Calculate the magnetization of the gas when it is in a magnetic field of \(1.2 \mathrm{~T}\).

A magnet in the form of a cylindrical rod has a length of \(5.00 \mathrm{~cm}\) and a diameter of \(1.00 \mathrm{~cm}\). It has a uniform magnetization of \(5.30 \times 10^{3} \mathrm{~A} / \mathrm{m}\). What is the magnitude of its magnetic dipole moment?
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