/*! 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 55 Interactive Solution \(\underlin... [FREE SOLUTION] | 91Ó°ÊÓ

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Interactive Solution \(\underline{22.55}\) at offers one approach to problems such as this one. The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 50: 1 . The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{~A} .\) Find the power consumed by the air filter.

Short Answer

Expert verified
The power consumed by the air filter is 10.2 W.

Step by step solution

01

- Understand the Problem

We have a step-up transformer with a turn ratio of 50:1. This means that for every turn on the primary coil, there are 50 turns on the secondary coil. The primary coil is connected to a 120V power source, and the current in the secondary coil is given as \(1.7 \times 10^{-3} \mathrm{~A}\). We need to find the power consumed by the air filter connected to the secondary coil.
02

- Calculate Secondary Voltage

Using the transformer's turns ratio, calculate the secondary voltage \(V_s\). The formula for the secondary voltage in terms of primary voltage \(V_p\) and turns ratio \(n\) is \(V_s = n \times V_p\). Here, \(n = 50\) and \(V_p = 120 \mathrm{~V}\). Substitute these values to get \(V_s = 50 \times 120\).
03

- Compute Secondary Voltage Result

Calculate the result from the secondary voltage formula: \(V_s = 50 \times 120 = 6000 \mathrm{~V}\). This is the voltage across the secondary coil.
04

- Calculate Power Consumed

The power consumed by the air filter can be calculated using the formula \(P = V_s \times I_s\), where \(I_s\) is the current in the secondary coil (\(1.7 \times 10^{-3} \mathrm{~A}\)). Substitute \(V_s = 6000\) into the formula to get \(P = 6000 \times 1.7 \times 10^{-3}\).
05

- Compute Power Result

Calculate the power using the substituted values: \(P = 6000 \times 1.7 \times 10^{-3} = 10.2 \mathrm{~W}\). This is the power consumed by the air filter.

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Key Concepts

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

Transformers Unveiled
In the world of electrical engineering, transformers are crucial devices that play a significant role in power distribution. Simply put, a transformer is a device that changes (or "transforms") the voltage of electrical power. They do this by transferring electrical energy between two or more circuits through electromagnetic induction. Transformers are essential when there's a need to increase or decrease voltage for efficient power transmission over long distances.
  • Electricity is either stepped up or stepped down depending on the need.
  • Transformers are equipped with primary and secondary coils (or windings).
  • The primary coil connects to the input voltage source, whereas the secondary coil connects to the output.
Let's dive a little deeper to understand the step-up transformer involved in the original exercise. A step-up transformer increases the voltage from the primary to the secondary coil. They are effectively used when higher voltage is required at the output without altering the overall power.
Turns Ratio Explained
One of the most critical components of a transformer is the turns ratio. 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)). Mathematically, this is expressed as:\[\text{Turns Ratio} = \frac{N_s}{N_p}\]This ratio is vital for determining the transformer's output voltage relative to its input voltage. In our step-up transformer example, a turns ratio of 50:1 indicates that for every single turn on the primary coil, the secondary coil has 50 turns.
  • A higher turns ratio means a higher voltage increase from primary to secondary.
  • The turns ratio directly influences calculations for secondary voltage.
Understanding the turns ratio helps in designing transformers for specific applications, ensuring they operate within desired parameters, like in our step-up situation where the voltage must be increased to operate a device such as an electrostatic air filter.
Understanding Secondary Voltage
The secondary voltage of a transformer is a key factor in its function. It represents the voltage that is output from the secondary coil. To calculate the secondary voltage, we often use the relation involving the turns ratio and the primary voltage (V_p):\[V_s = \left(\frac{N_s}{N_p}\right) \times V_p\]In the resolved exercise, the primary voltage (V_p) is 120 V and the turns ratio is 50, giving us a secondary voltage of 6000 V.
  • The secondary voltage supports the operational requirements of devices connected to the transformer.
  • It is essential for ensuring enough voltage is available for end applications, like powering an air filter.
Calculating the secondary voltage is a fundamental step in verifying a transformer’s design and ensuring that it meets the required electrical characteristics.
Calculating Electric Power
Power consumption in electrical devices can be calculated using a straightforward formula involving voltage and current:\[P = V \times I\]Where \(P\) is the power, \(V\) is the voltage, and \(I\) is the current. For the air filter connected to the secondary coil of our transformer, this means using the previously calculated secondary voltage and the secondary current given as \(1.7 \times 10^{-3} \mathrm{~A}\). Thus, the power is computed as follows:\[P = 6000 \times 1.7 \times 10^{-3} = 10.2 \mathrm{~W}\]Understanding power calculation is crucial. It guides engineers and technicians in designing electrical systems that operate safely and efficiently:
  • It provides a direct measure of the energy consumed over time, influencing cost and efficiency.
  • Knowledge of power consumption helps in selecting appropriate components for electrical setups, ensuring that they are neither underpowered nor overburdened.
In practice, accurate power calculations like this ensure devices function correctly and contribute to the system's effectiveness.

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Most popular questions from this chapter

Two coils of wire are placed close together. Initially, a current of 2.5 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2} \mathrm{~s}\). During this time, the average emf induced in the other coil is \(1.7 \mathrm{~V}\). What is the mutual inductance of the two-coil system?

A solenoid has a cross-sectional area of \(6.0 \times 10^{-4} \mathrm{~m}^{2},\) consists of 400 turns per meter, and carries a current of 0.40 A. A 10 -turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a \(1.5-\Omega\) resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of \(0.050 \mathrm{~s} .\) Find the average current induced in the coil.

The armature of an electric drill motor has a resistance of \(15.0 \Omega\). When connected to a 120.0-V outlet, the motor rotates at its normal speed and develops a back emf of \(108 \mathrm{~V}\). (a) What is the current through the motor? (b) If the armature freezes up due to a lack of lubrication in the bearings and can no longer rotate, what is the current in the stationary armature? (c) What is the current when the motor runs at only half speed?

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?

Concept Questions The drawing shows a straight wire carrying a current \(I\). Above the wire is a rectangular loop that contains a resistor \(R\). (a) Does the magnetic field produced by the current \(I\) penetrate the loop and generate a magnetic flux? (b) When is there an induced current in the loop, if the current \(I\) is constant or if it is decreasing in time? (c) When there is an induced magnetic field produced by the loop, does it always have a direction that is opposite to the direction of the magnetic field produced by the current \(I\) ? Provide a reason for each answer. Problem If the current \(I\) is decreasing in time, what is the direction of the induced current through the resistor \(R\) - left to right or right to left? Give your reasoning.

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