/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A Step-Down Transformer. A trans... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(12.0 \mathrm{~V}\) (rms) to a portable electronic device. The load resistance in the secondary is \(5.00 \Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply?

Short Answer

Expert verified
(a) The turns ratio is 10. (b) The secondary rms current is 2.4 A.

Step by step solution

01

Understanding the Step-Down Transformer

A step-down transformer decreases voltage from primary to secondary coil. For this, the ratio of the number of turns of wire in the primary coil (\(N_p\)) to the number of turns in the secondary coil (\(N_s\)) dictates the voltage transformation.
02

Using the Transformer Equation (Part a)

The formula for a transformer is \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \), where \(V_p\) and \(V_s\) are the primary and secondary voltages, respectively. For a step-down transformer supplying 12V from 120V, the turns ratio is \( \frac{N_p}{N_s} = \frac{120}{12} \).
03

Calculating the Turns Ratio

Substitute the given voltages into the transformer equation to find the ratio: \( \frac{120}{12} = 10 \). Hence, the turns ratio of primary to secondary coils, \( \frac{N_p}{N_s} \), is 10.
04

Determining the Rms Current in the Secondary (Part b)

Ohm's Law \( V = IR \) is used in this step. Given that the secondary voltage \( V_s = 12 \, \text{V} \) and resistance \( R = 5.00 \, \Omega \), solve for current \( I_s \): \( I_s = \frac{V_s}{R} = \frac{12}{5.00} \).
05

Calculating the Rms Current

Compute the current \(I_s\) using the values from Ohm's Law: \( I_s = \frac{12}{5.00} = 2.4 \, \text{A} \). Hence, the secondary rms current required is 2.4 A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Step-Down Transformer
A step-down transformer is an electrical device designed to reduce voltage from a higher level in the primary coil to a lower level in the secondary coil. It is an essential component in many electronic devices, especially where lower voltage is needed to operate safely and efficiently. The transformer works by using electromagnetic induction to transfer electrical energy between two circuits.
  • The primary coil is the input side, connected to the higher voltage source.
  • The secondary coil is the output side, delivering the lowered voltage.
The functionality of a step-down transformer is crucial for converting domestic AC voltage, such as the common 120V in a wall outlet, to a more manageable level for smaller electronics like portable devices. This reduction in voltage also helps in minimizing the risk of electrical hazards by ensuring devices operate within their safe voltage range.
Turns Ratio
The turns ratio is a fundamental concept in transformer operation that dictates how voltages are transformed between the primary and secondary coils. It is represented by the ratio of the number of turns in the primary coil (N_p) to the number of turns in the secondary coil (N_s).
The turns ratio can be calculated using the voltage ratio formula:
  • \[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]
  • In a step-down transformer, if the primary voltage (V_p) is 120V and the secondary voltage (V_s) is 12V, the turns ratio would be:\( \frac{120}{12} = 10 \).
This means that the primary coil has 10 times more turns than the secondary coil. By adjusting this ratio, transformers can efficiently convert voltage levels to meet the needs of various electronic applications.
RMS Current
RMS, or root mean square current, is an important measurement in AC circuits that provides the equivalent value of a DC current. It represents the effective current that actually performs work in the circuit. For a transformer, the rms current is especially important in evaluating the load the transformer can handle in its secondary coil.
To determine the secondary rms current, Ohm’s Law is applied:
  • \( V = IR \),
  • where \( I_s \) is the rms current, \( V_s = 12 \, \text{V} \) and the load resistance \( R = 5.00 \, \Omega \), results in:\( I_s = \frac{12}{5.00} = 2.4 \, \text{A} \).
Thus, the secondary coil supplies an rms current of 2.4 A, ensuring that the secondary load is powered properly without exceeding the transformer’s operational limits.
Ohm's Law
Ohm's Law is a foundational principle used extensively to analyze electrical circuits, especially in understanding transformers. It establishes a relationship between voltage (V), current (I), and resistance (R), expressed as:
  • \( V = IR \)
  • This equation helps calculate how much current flows through a circuit when a certain voltage is applied across a resistance.
In the context of a transformer’s secondary side, Ohm’s Law is used to determine the amount of rms current supplied to the load. For example, with a secondary voltage \( V_s = 12 \, \text{V} \) and resistance \( R = 5.00 \, \Omega \), the rms current \( I_s \) is found by rearranging the formula to \( I_s = \frac{V_s}{R} \).
Applying Ohm's Law in transformer calculations not only ensures proper design and functionality but also guarantees safety by preventing overloading.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At a frequency \(\omega_{1}\) the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to \(\omega_{2}=2 \omega_{1}\), what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is changed to \(\omega_{3}=\omega_{1} / 3\), what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance \(R\) to form an \(L-R-C\) series circuit, what will be the resonance angular frequency of the circuit?

An LCR series circuit with \(100 \Omega\) resistance is connected to an \(A C\) source of \(200 V\) and angular frequency 300 radians/sec. When only the capacitance is removed The current lags behind the voltage by \(60^{\circ} .\) When only the inductance is removed the current leads the voltage \(60^{\circ}\). Calculate the current and the power dissipated in the LCR circuit.

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{~V}, R=80.0 \Omega\), and the reactance of the capacitor is \(480 \Omega .\) The voltage amplitude across the capacitor is \(360 \mathrm{~V}\). (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

A series circuit has an impedance of \(60.0 \Omega\) and a power factor of \(0.8\) at \(50.0 \mathrm{~Hz}\). The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

You have a \(200-\Omega\) resistor, a \(0.400-H\) inductor, and a \(6.00-\mu \mathrm{F}\) capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude \(30.0 \mathrm{~V}\) and an angular frequency of \(250 \mathrm{rad} / \mathrm{s}\). (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle \(\phi\) of the source voltage with respect to the current? Does the source voltage lag or lead the current? (e) Construct the phasor diagram.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.