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Chapter 22: Problem 60

The battery charger for an MP3 player contains a step-down transformer with a turns ratio of \(1: 32,\) so that the voltage of \(120 \mathrm{V}\) available at a wall socket can be used to charge the battery pack or operate the player. What voltage does the secondary coil of the transformer provide?

Short Answer

Expert verified
The secondary coil provides 3.75 V.

Step by step solution

01

Identify the Known Values

We are given the turns ratio of the transformer, which is \( 1:32 \). This means the primary coil has 1 turn for every 32 turns in the secondary coil. Also, the primary voltage \( V_p \) is provided as \( 120 \text{ V} \). We need to find the secondary voltage \( V_s \).
02

Use the Transformer Voltage Ratio Formula

The voltage ratio in transformers is given by the formula: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \] where \( V_s \) and \( V_p \) are the secondary and primary voltages, respectively, and \( N_s \) and \( N_p \) are the number of turns in the secondary and primary coils, respectively. Here, \( \frac{N_s}{N_p} = 32 \) and \( V_p = 120 \text{ V} \).
03

Solve for the Secondary Voltage \(V_s\)

Rearrange the formula to solve for \( V_s \): \[ V_s = V_p \times \frac{N_s}{N_p} \] Substitute the known values: \( V_s = 120 \times 32 \). Calculate \( V_s \).
04

Calculate the Value

Perform the calculation: \( V_s = 120 \times \frac{1}{32} = \frac{120}{32} \). Simplifying gives \( V_s = 3.75 \text{ V} \). This is the voltage provided by the secondary coil.

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Key Concepts

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

Understanding Turns Ratio
The turns ratio in a transformer is a vital concept that defines how voltages are transformed between two circuits. It is expressed as the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. In this exercise, the turns ratio is given as \(1:32\).
This ratio tells us that for every single turn in the primary coil, the secondary coil has 32 turns. The turns ratio directly influences the voltage transformation process in transformers. By understanding this ratio, you can predict how the voltage will increase or decrease as it moves from one coil to the other.
Exploring Primary Voltage
The primary voltage refers to the voltage supplied to the primary winding of the transformer. In our scenario, the primary voltage \( V_p \) is \( 120 \text{ V} \).
This is the voltage that comes from an external source, like a power outlet, before being stepped down to a lower voltage through the transformer.
  • The primary voltage is crucial because it dictates how much electrical energy enters the transformer.
  • In our example, the 120V power is standard for residential electrical outlets, making it easy and convenient to use in small household devices.
The role of the primary voltage is to provide the necessary input for transformation and dictate the potential energy available for changes within a system.
Calculating Secondary Voltage
The secondary voltage is the voltage that comes out of the secondary winding of the transformer. To find the secondary voltage \( V_s \), we use the transformer voltage ratio formula:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]
Here, \( \frac{N_s}{N_p} = 32 \), so we rearrange the formula to solve for \( V_s \):
\[ V_s = V_p \times \frac{N_s}{N_p} \]
In our example, substituting the known values gives:
\[ V_s = 120 \times \frac{1}{32} = \frac{120}{32} = 3.75 \text{ V} \]
  • The secondary voltage is the one used to power the secondary circuit.
  • In this example, it鈥檚 reduced from 120V to a much safer 3.75V, appropriate for charging small devices such as an MP3 player.
Function of a Step-Down Transformer
A step-down transformer is specifically designed to decrease voltage. In our exercise, the transformer reduces the wall voltage from 120V down to 3.75V.
Step-down transformers are widely used in everyday household electronics to make sure devices receive the correct voltage needed for operation. By lowering voltage, they make it safer and compatible with smaller electronic devices.
  • They make high-voltage power from outlets usable for low-voltage devices.
  • Step-down transformers prevent damage to electronics that require lower voltages.
  • They are essential for modern electrical systems, ensuring devices can function properly without being exposed to dangerous high voltages.

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Most popular questions from this chapter

A flat circular coil with 105 turns, a radius of \(4.00 \times 10^{-2} \mathrm{m}\), and a resistance of \(0.480 \Omega\) is exposed to an external magnetic field that is directed perpendicular to the plane of the coil. The magnitude of the external magnetic field is changing at a rate of \(\Delta B / \Delta t=0.783 \mathrm{T} / \mathrm{s}\), thereby inducing a current in the coil. Find the magnitude of the magnetic field at the center of the coil that is produced by the induced current.

A piece of copper wire is formed into a single circular loop of radius \(12 \mathrm{cm} .\) A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to \(0.60 \mathrm{T}\) in a time of \(0.45 \mathrm{s}\). The wire has a resistance per unit length of \(3.3 \times 10^{-2} \Omega / \mathrm{m} .\) What is the average electrical energy dissipated in the resistance of the wire?

A magnetic field is passing through a loop of wire whose area is \(0.018 \mathrm{m}^{2} .\) The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of \(0.20 \mathrm{T} / \mathrm{s}\) (a) Determine the magnitude of the emf induced in the loop. (b) Suppose that the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) should the area be changed at the instant when \(B=1.8 \mathrm{T}\) if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.

During a 72 -ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a \(6.0-\mathrm{mA}\) 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 mH. What is the change in the primary current?

A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of 1.7 'I. An emf that has a magnitude of \(2.6 \mathrm{V}\) is induced in this coil because the coil's area \(A\) is shrinking. What is the magnitude of \(\Delta A / \Delta t,\) which is the rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) at which the area changes?
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