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Use Phasors Add the quantities $$ \begin{array}{l} y_{1}=10 \sin \omega t \\ y_{2}=15 \sin \left(\omega t+30^{\circ}\right) \\ y_{3}=5.0 \sin \left(\omega t-45^{\circ}\right) \end{array} $$ using the phasor method.

Short Answer

Expert verified
The resulting sinusoidal function is \(y = 26.83 \, \text{sin} \, (\text{ωt} + 8.50^\text{°})\).

Step by step solution

01

- Represent each sinusoidal function as a phasor

Convert each of the sinusoidal functions into their respective phasor representations. For a sinusoidal function of the form \(y = A \, \text{sin}(\theta)\), the phasor representation is \(A \, \text{∠} θ\). Thus, we have:\(y_1 = 10 \, \text{sin} (\text{ωt})\) becomes \(10 \, \text{∠} \, 0^\text{°}\),\(y_2 = 15 \, \text{sin} (\text{ωt} + 30^\text{°})\) becomes \(15 \, \text{∠} \, 30^\text{°}\),\(y_3 = 5.0 \, \text{sin} (\text{ωt} - 45^\text{°})\) becomes \(5.0 \, \text{∠} \, -45^\text{°}\).
02

- Convert phasors to rectangular form

To add the phasors, convert them into rectangular form using the equations:\[ A \, \text{∠} \, \theta = A \, (\text{cos} \, \theta + j \, \text{sin} \, \theta) \].So, we have:\(10 \, \text{∠} \, 0^\text{°} = 10 \, (\text{cos} \, 0^\text{°} + j \, \text{sin} \, 0^\text{°}) = 10 + 0j\),\(15 \, \text{∠} \, 30^\text{°} = 15 \, (\text{cos} \, 30^\text{°} + j \, \text{sin} \, 30^\text{°}) = 15 \, (\frac{\text{√3}}{2} + j \, \frac{1}{2}) = 12.99 + 7.5j\),\(5.0 \, \text{∠} \, -45^\text{°} = 5.0 \, (\text{cos} \, (-45^\text{°}) + j \, \text{sin} \, (-45^\text{°})) = 5.0 \, (\frac{1}{\text{√2}} + j \, (-\frac{1}{\text{√2}})) = 3.54 - 3.54j\).
03

- Add the phasors in rectangular form

Sum the real and imaginary parts of all the phasors separately.Real part: \(10 + 12.99 + 3.54 = 26.53\),Imaginary part: \(0 + 7.5 - 3.54 = 3.96j\).Thus, the combined phasor in rectangular form is \(26.53 + 3.96j\).
04

- Convert the resultant phasor back to polar form

Convert the resulting rectangular form phasor back to polar form. Use the equations:\[ R = \text{√} ( \text{Re}^2 + \text{Im}^2 ) \],\[ θ = \text{tan}^{-1} \, (\frac{\text{Im}}{\text{Re}}) \].First, calculate the magnitude (\(R\)):\(R = \text{√} (26.53^2 + 3.96^2) = 26.83\).Next, calculate the angle (\(θ\)):\(θ = \text{tan}^{-1} \, (\frac{3.96}{26.53}) = 8.50^\text{°}\).So the resultant phasor is \(26.83 \, \text{∠} \, 8.50^\text{°}\).
05

- Write the resulting sinusoidal function

Finally, convert the resulting phasor back to a sinusoidal function. The phasor \(26.83 \, \text{∠} \, 8.50^\text{°}\) corresponds to:\[ y = 26.83 \, \text{sin} \, (\text{ωt} + 8.50^\text{°}) \].Therefore, the resulting sinusoidal function is \(y = 26.83 \, \text{sin} \, (\text{ωt} + 8.50^\text{°})\).

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Key Concepts

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

Sinusoidal Functions
A sinusoidal function is a mathematical function that describes a smooth periodic oscillation. It can be in the form of a sine or cosine function. In engineering and physics, sinusoidal functions are essential for describing alternating current (AC) waveforms.

The general representation of a sinusoidal function is: \[ y = A \, \text{sin} \, (\theta) \] where:
  • \( A \) is the amplitude, which represents the peak value of the oscillation.
  • \( \theta \) is the phase angle, which represents the initial angle at time zero.
  • 'sin' (sine) determines the shape of the waveform.
In the context of AC circuits, it is common to use sinusoidal functions to denote voltages and currents, which alternate between positive and negative values over time.

For the given exercise, we used sinusoidal functions to represent voltages or currents at various phases.
Rectangular Form
Rectangular form is a way to represent complex numbers, which can be especially useful in electrical engineering when dealing with sinusoidal functions. This form represents a complex number as: \[ z = x + jy \]

Here:
  • \( x \) is the real part.
  • \( y \) is the imaginary part.
  • \( j \) is the imaginary unit, equivalent to \( \text{√} {-1} \).
Using the rectangular form is particularly advantageous when adding or subtracting complex numbers.

For the given exercise, to add sinusoidal functions, we converted phasors to rectangular form using the formula: \[ A \, \text{∠} \, \theta = A \, (\text{cos} \, \theta + j \, \text{sin} \, \theta) \] This conversion allowed us to combine the real and imaginary parts separately, thereby simplifying the addition process.
Polar Form
Polar form represents a complex number in terms of its magnitude and angle. This is highly useful in AC circuit analysis and when working with phasors. A complex number in polar form is written as: \[ z = R \, \text{∠} \, \theta \]

Here:
  • \( R \) is the magnitude and can be calculated as \( R = \text{√} (x^2 + y^2) \).
  • \( \theta \) is the phase angle and can be calculated as \( \theta = \text{tan}^{-1} \, (y/x) \).
The magnitude represents the length of the vector in the complex plane, while the phase angle represents the orientation of the vector.

In the exercise, converting the resultant rectangular form phasor to polar form involved calculating these magnitude and angle values to write the final phasor representation.
Phasor Representation
Phasor representation is a way to simplify the analysis of AC circuits by converting sinusoidal functions to complex numbers. Phasors translate the time domain problem to the frequency domain, making the math easier.

A phasor is generally represented as: \[ A \, \text{∠} \, \theta \] where:
  • \( A \) is the amplitude.
  • \( \theta \) is the phase angle.
Phasors make use of Euler's formula, \( e^{j\theta} = \text{cos} \, \theta + j \, \text{sin} \, \theta \), to transform sinusoidal functions into complex exponential form.

To solve the given problem, we first converted sinusoidal functions to phasors and then manipulated them using the phasor method. This method simplifies the addition, subtraction, and multiplication of sinusoidal functions, making it a crucial tool in electrical engineering.

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Most popular questions from this chapter

Airtight Chamber In Fig. 36-28, an airtight chamber \(5.0 \mathrm{~cm}\) long with glass windows is placed in one arm of a Michelson interferometer. Light of wavelength \(\lambda=500 \mathrm{~nm}\) is used. Evacuating the air from the chamber causes a shift of 60 fringes. From these data, find the index of refraction of air at atmospheric pressure.

Angular Separation In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen \(50.0 \mathrm{~cm}\) from the slits?

Two Glass Plates Two glass plates are held together at one end to form a wedge of air that acts as a thin film. A broad beam of light of wavelength \(480 \mathrm{~nm}\) is directed through the plates, perpendicular to the first plate. An observer intercepting light reflected from the plates sees on the plates an interference pattern that is due to the wedge of air. How much thicker is the wedge at the sixteenth bright fringe than it is at the sixth bright fringe, counting from where the plates touch?

Directly Downward In Fig. \(36-25\), white light is sent directly downward through the top plate of a pair of glass plates. The plates touch at the left end and are separated by a wire of diameter \(0.048\) \(\mathrm{mm}\) at the right end; the air between the plates acts as a thin film. An observer looking down through the top plate sees bright and dark fringes due to that film. (a) Is a dark fringe or a bright fringe seen at the left end? (b) To the right of that end, fully destructive interference occurs at different locations for different wavelengths of the light. Does it occur first for the red end or the blue end of the visible spectrum?

Soap Film Light of wavelength \(624 \mathrm{~nm}\) is incident perpendicularly on a soap film (with \(n=1.33\) ) suspended in air. What are the least two thicknesses of the film for which the reflections from the film undergo fully constructive interference?

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