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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

A rectangular loop of wire with sides 0.20 and \(0.35 \mathrm{~m}\) lies in a plane perpendicular to a constant magnetic field (see part \(a\) of the drawing). The magnetic field has a magnitude of \(0.65 \mathrm{~T}\) and is directed parallel to the normal of the loop's surface. In a time of \(0.18 \mathrm{~s}\), one-half of the loop is then folded back onto the other half, as indicated in part \(b\) of the drawing. Determine the magnitude of the average emf induced in the loop.

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume your house is the only one using electric power. At a substation the primary of a step- down transformer (turns ratio \(=1: 29\) ) receives the voltage from the high- voltage transmission line. Because of your usage, a current of \(48 \mathrm{~mA}\) exists in the primary of this transformer. The secondary is connected to the primary of another step-down transformer (turns ratio \(=1: 32\) ) somewhere near your house, perhaps up on a telephone pole. The secondary of this transformer delivers a 240-V emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.

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 \(1.7 \mathrm{~T}\). 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 \(\left(\right.\) in \(\left.m^{2} / s\right)\) at which the area changes?

The drawing shows a coil of copper wire that consists of two semicircles joined by straight sections of wire. In part \(a\) the coil is lying flat on a horizontal surface. The dashed line also lies in the plane of the horizontal surface. Starting from the orientation in part \(a\), the smaller semicircle rotates at an angular frequency \(\omega\) about the dashed line, until its plane becomes perpendicular to the horizontal surface, as shown in part \(b\). A uniform magnetic field is constant in time and is directed upward, perpendicular to the horizontal surface. The field completely fills the region occupied by the coil in either part of the drawing. (a) In which part of the drawing, if either, does a greater magnetic flux pass through the coil? Account for your answer. (b) As the shape of the coil changes from that in part \(a\) of the drawing to that in part \(b\), does an induced current flow in the coil, and, if so, in which direction does it flow? Give your reasoning. To describe the flow, imagine that you are above the coil looking down at it. (c) How is the period \(T\) of the rotational motion related to the angular frequency \(\omega\), and in terms of the period, what is the shortest time interval that elapses between parts \(a\) and \(b\) of the drawing? the magnitude of the magnetic field is \(0.35 \mathrm{~T}\). The resistance of the coil is \(0.025 \Omega\), and the smaller semicircle has a radius of \(0.20 \mathrm{~m}\). The angular frequency at which the small semicircle rotates is \(1.5 \mathrm{rad} / \mathrm{s}\). Determine the average current, if any, induced in the coil as the coil changes shape from that in part \(a\) of the drawing to that in \(\operatorname{part} b\).

A circular coil \((950\) turns, radius \(=0.060 \mathrm{~m})\) is rotating in a uniform magnetic field. At \(t=0 \mathrm{~s}\), the normal to the coil is per pendicular to the magnetic field. At \(t=0.010 \mathrm{~s}\) the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made oneeighth of a revolution. An average emf of magnitude \(0.065 \mathrm{~V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

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