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Chapter 22: Problem 60

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.

Short Answer

Expert verified
Your house uses 368.64 watts of power.

Step by step solution

01

Understand Primary Transformer Current and Voltage

We have a primary current of 48 mA (rms) in the first step-down transformer. The turns ratio of this transformer is 1:29, meaning the secondary voltage is reduced by a factor of 29 compared to the primary voltage from the high-voltage transmission line.
02

Calculate Primary Voltage of the First Transformer

Since we need the power used, and power is given by \( P = I \times V \), where \( I \) is the current and \( V \) is the voltage, we start by determining the voltage on the primary side of the first transformer. However, without directly knowing this primary voltage, we can express it in terms of our known variables. We know that the voltage of the secondary of the second transformer is 240 V.
03

Determine Voltage on the Secondary of First Transformer

The voltage on the secondary of the first transformer is the primary voltage of the second transformer which experiences the second transformer's turns ratio to reach the 240 V delivered to the house. Let \( V_2 \) be the secondary voltage of the first transformer, then: \[ V_2 = 32 \times 240 \text{ V}\]Calculating gives \( V_2 = 7680 \text{ V} \).
04

Calculate Primary Power

The power delivered from the primary of the first transformer to the secondary (which is the same as the second transformer's primary), is calculated by the power formula: \[ P = I_{primary,1} \times V_{secondary,1}\]Substitute the known values: \[ P = 48 \times 10^{-3} \text{ A} \times 7680 \text{ V}\]Calculate to find: \[ P = 368.64 \text{ W}\]
05

Verify the Power Consumed by the House

Since transformers ideally do not lose power (assuming 100% efficiency in this problem), all the power calculated at the secondary of the first transformer ultimately is delivered as power consumed by your house. Thus, the house uses 368.64 W.

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Key Concepts

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

Step-down Transformer
A step-down transformer is a device that reduces high voltage to a lower voltage. This makes it useful in power distribution systems where electricity leaves power stations at high voltages.
By adjusting the voltage to safer, lower levels, it can be used in residential areas.
A transformer consists of two coils of wire called the primary and secondary windings. The ratio of the number of turns in these coils is known as the turns ratio. For a step-down transformer, the turns ratio is such that the primary coil has more turns than the secondary.
Key features of step-down transformers include:
  • Reduction of voltage: The turns ratio directly influences how much the voltage decreases.
  • Maintaining power: Ideally, transformers retain the power into consideration that some energy might be lost as heat.
In our example, the primary coil of the first transformer has a turns ratio of 1:29, meaning the secondary voltage will be 1/29th of the primary voltage.
Similarly, the second transformer near the house further steps down the voltage with a turns ratio of 1:32.
Power Formula
The power formula is a crucial part of understanding electrical systems. It allows you to calculate how much work an electrical circuit can accomplish.
The power formula is given by:\[P = I \times V\] where:
  • \( P \) is power measured in watts (W).
  • \( I \) is the current in amperes (A).
  • \( V \) is the voltage in volts (V).
Power calculation is vital for determining electricity consumption, like in the problem when analyzing the power used by the house.
In our situation, the primary of the first transformer experiences a current of 48 mA and a calculated secondary voltage of 7680 V before continuing on to supply power to the house.
Using the power formula, we calculated the power output to be 368.64 W, showing how efficiently energy gets distributed.
RMS Values
RMS, or Root Mean Square, values make working with alternating current (AC) a lot easier. They provide a means to describe the equivalent direct current (DC) value, making calculations manageable.
RMS voltage or current reflects the effective value of AC voltage or current. Here's why RMS values matter:
  • They are used in power calculations for AC circuits to find the same value as if the system were a DC circuit.
  • They help standardize and measure quantities despite the constant change in AC.
In the exercise, both the current and the voltage values are given as RMS.
This is crucial because, without using RMS values, our power calculation using the standard power formula wouldn't accurately represent the true power transferred to your house.
Hence, RMS ensures that the values are equipped to calculate real-world power scenarios effectively.

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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.

During a 72 -ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a \(6.0\) -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 \(3.2 \mathrm{mH}\). What is the change in the primary current?

A \(3.0-\mu\) F capacitor has a voltage of 35 V between its plates. What must be the current in a 5.0 -mH inductor, such that the energy stored in the inductor equals the energy stored in the capacitor?

ssm www The drawing shows a plot of the output emf of a generator as a function of time \(t\). The coil of this device has a cross-sectional area per turn of \(0.020 \mathrm{~m}^{2}\) and contains 150 turns. Find (a) the frequency \(f\) of the generator in hertz, (b) the angular speed \(\omega\) in \(\mathrm{rad} / \mathrm{s}\), and \((\mathrm{c})\) the magnitude of the magnetic field

A generating station is producing \(1.2 \times 10^{6} \mathrm{~W}\) of power that is to be sent to a small town located \(7.0 \mathrm{~km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of length of \(5.0 \times 10^{-2} \Omega / \mathrm{km} .\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{~V} .\) (b) A \(100: 1\) step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?
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