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Chapter 5: Problem 11

The typical voltage, \(V\), in volts (V), supplied by an electrical outlet in Cuba is a sinusoidal function that oscillates between \(-155 \mathrm{V}\) and \(+155 \mathrm{V}\) and makes 60 complete cycles each second. Determine an equation for the voltage as a function of time, \(t.\)

Short Answer

Expert verified
The equation is \( V(t) = 155 \sin(120\pi t) \).

Step by step solution

01

Identify the amplitude

The amplitude \(A\) of the sinusoidal function is the maximum voltage, which is given as \(155 \text{ volts}\).
02

Determine the period

The period \(T\) is the time taken for one complete cycle. Since there are 60 cycles per second, \(T = \frac{1}{60} \text{ seconds}\).
03

Calculate the angular frequency

The angular frequency \(\omega\) is given by \(\omega = \frac{2\pi}{T}\). Substituting the value of \(T\) from Step 2, we get \(\omega = 2\pi \cdot 60 = 120\pi \text{ radians per second}\).
04

Write the general equation

A sinusoidal function can be represented as \( V(t) = A \sin(\omega t) \), where \(A\) is the amplitude and \(\omega\) is the angular frequency.
05

Substitute the values

Using the values determined: \( A = 155 \) and \( \omega = 120\pi \), the equation becomes \[ V(t) = 155 \sin(120\pi t) \].

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Key Concepts

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

Amplitude
In a sinusoidal function, the amplitude represents the peak value of the oscillation. It is essentially the maximum absolute value it reaches from its mean value. In our electrical engineering example, the voltage oscillates between \(+155 V\) and \(-155 V\). Thus, the amplitude \(A\) is \(155 V\). This means that the highest voltage the sinusoidal function will reach is \(+155 V\), and the lowest will be \(-155 V\). The amplitude represents half the total range of the oscillation (difference between the maximum and minimum values), clarifying the strength or intensity of the voltage fluctuation.
Period
The period \(T\) of a sinusoidal function is the time it takes for one complete cycle of the wave to occur. In this case, it's the duration for which the voltage oscillation starts from \(0 V\), reaches \(+155 V\), goes back to \(-155 V\), and then returns to \0 V\. With 60 cycles occurring every second, each cycle must have a time period of \(\frac{1}{60}\) seconds. This means that every fraction of \(\frac{1}{60}\) seconds, the voltage pattern repeats itself, ensuring the waveform is consistent over time.
Angular Frequency
Angular frequency \( \omega \) is a measure of how quickly the sinusoidal function oscillates and is related to the period. It's given by the formula \(\omega = \frac{2\pi}{T} \), where \T\ is the period. For the voltage with a period of \( \frac{1}{60} \) seconds, substituting in the formula yields \(\omega = \frac{2\pi}{\frac{1}{60}} = 120\pi \) radians per second. This value indicates how fast the oscillations are in terms of angular motion, representing how many radians the function goes through per second.
Sinusoidal Function
A sinusoidal function describes a smooth, repetitive oscillation and is fundamental in electrical engineering to explain alternating current (AC). The general equation of a sinusoidal function is \( V(t) = A \sin(\omega t) \), where \A\ is the amplitude, and \ \omega\ is the angular frequency. Using the amplitude \A\ of \(155 V\) and angular frequency \ \omega\ of \120\ \pi\ radians per second, we derive the specific function for the voltage in the given problem: \( V(t) = 155 \sin(120\pi t) \). This equation succinctly combines both the peak voltage and the frequency of the oscillation to describe the voltage as a function of time.

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Most popular questions from this chapter

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