Problem 54 - Calculus

Question

Household electric current can be modeled by the voltage \(V=\hat{V} \sin (120 \pi t+\phi),\) where \(t\) is measured in seconds, \(\hat{V}\) is the maximum value that \(V\) can attain, and \(\phi\) is the phase angle. Such a voltage is usually said to be 60 -cycle, since in 1 second the voltage goes through 60 oscillations. The root-mean-square voltage, usually denoted by \(V_{\mathrm{rms}}\) is defined to be the square root of the average of \(V^{2} .\) Hence $$ V_{\mathrm{rms}}=\sqrt{\int_{\phi}^{1+\phi}(\hat{V} \sin (120 \pi t+\phi))^{2} d t} $$ A good measure of how much heat a given voltage can produce is given by \(V_{\mathrm{rms}}\) (a) Compute the average voltage over 1 second. (b) Compute the average voltage over \(1 / 60\) of a second. (c) Show that \(V_{\mathrm{rms}}=\frac{\hat{V} \sqrt{2}}{2}\) by computing the integral for \(V_{\mathrm{rms}}\) Hint: \(\int \sin ^{2} t d t=-\frac{1}{2} \cos t \sin t+\frac{1}{2} t+C\) (d) If the \(V_{\mathrm{rms}}\) for household current is usually 120 volts, what is the value \(\hat{V}\) in this case?

Step-by-Step Solution

Answer

(a) 0 volts, (b) 0 volts, (c) \(V_{\mathrm{rms}} = \frac{\hat{V}\sqrt{2}}{2}\), (d) \(\hat{V} \approx 169.7\) volts.

Step 1: Compute Average Voltage Over 1 Second

The average voltage over a period is computed as follows:\[ V_{avg} = \frac{1}{T} \int_0^T V(t) \, dt \]For part (a), the period \(T\) is 1 second. Substituting the given voltage function, we have:\[ V_{avg} = \frac{1}{1} \int_0^1 \hat{V} \sin(120\pi t + \phi) \, dt \]The integral of a sinusoidal function over its period is zero, so the average voltage is:\[ V_{avg} = 0 \].

Step 2: Compute Average Voltage Over 1/60 of a Second

For part (b), the period \( T \) is \( \frac{1}{60} \) seconds:\[ V_{avg} = \frac{1}{1/60} \int_0^{1/60} \hat{V} \sin(120\pi t + \phi) \, dt \]This replicates the same logic since the integral of a sinusoidal function from 0 to \( \frac{1}{60} \) over its period still results in zero due to symmetry:\[ V_{avg} = 0 \].

Step 3: Show that \( V_{\mathrm{rms}} = \frac{\hat{V} \sqrt{2}}{2} \)

(c) We are given the expression of \( V_{\mathrm{rms}} \):\[ V_{\mathrm{rms}} = \sqrt{\int_{\phi}^{1+\phi} (\hat{V} \sin(120\pi t + \phi))^2 \, dt} \]This simplifies to:\[ V_{\mathrm{rms}} = \sqrt{\hat{V}^2 \int_{\phi}^{1+\phi} \sin^2(120\pi t + \phi) \, dt} \]Perform the integration using the hint:\[ \int \sin^2 t \, dt = -\frac{1}{2} \cos t \sin t + \frac{1}{2} t + C \]For one complete cycle:\[ \int_{\phi}^{\pi + \phi} \sin^2(x) \, dx = \frac{1}{2} \left( t \right) \Big|_{\phi}^{1 + \phi} = \frac{1}{2} (1) \]Thus, we have:\[ V_{\mathrm{rms}} = \sqrt{\hat{V}^2 \cdot \frac{1}{2}} = \frac{\hat{V}}{\sqrt{2}} \]This should be multiplied by \( \sqrt{2} \) to yield:\[ V_{\mathrm{rms}} = \frac{\hat{V} \sqrt{2}}{2} \].

Step 4: Find \( \hat{V} \) When \( V_{\mathrm{rms}} = 120 \) Volts

(d) Given \( V_{\mathrm{rms}} = 120 \) volts, find \( \hat{V} \):\[ 120 = \frac{\hat{V} \sqrt{2}}{2} \]Solving for \( \hat{V} \):\[ \hat{V} = 120 \times \frac{2}{\sqrt{2}} = 120 \sqrt{2} \approx 169.7 \]

Key Concepts

Sinusoidal Function

Sinusoidal functions are widely used in mathematics and physics to model periodic phenomena, such as waves and oscillations. These functions are characterized by their repetitive pattern, resembling the shape of a sine or cosine wave.
Here's what you need to know:

• Sine Function: The general form of a sine function is expressed as \( y = A \sin(Bx + C) + D \), where:
  • \( A \) determines the amplitude, which is the peak height of the wave from its central axis.
  • \( B \) affects the period, the length of one complete cycle of the wave, calculated as \( \frac{2\pi}{B} \).
  • \( C \) represents the phase shift, modifying where the wave starts along the x-axis.
  • \( D \) is the vertical shift, moving the wave up or down from the center.
• Periodicity: A key feature is that sinusoids repeat indefinitely every period, making them useful for modeling scenarios like alternating current (AC) in an electrical circuit.
• Applications: Beyond AC circuits, sinusoidal functions find applications in sound waves, tidal movements, and even the oscillation of springs.

If you remember these defining elements, you'll see sinusoids everywhere in both mathematical problems and real-world examples.

Phase Angle

The phase angle, typically denoted as \( \phi \), plays a critical role in sinusoidal functions, determining where the wave begins along the horizontal axis. Consider these points:

• Definition: At its core, the phase angle is the initial angle of a sinusoidal function. It influences the starting point of the wave's cycle, effectively shifting it forward or backward along the horizontal axis.
• Influence on Graph: By altering the phase angle:
  • You change the alignment of the peaks and troughs with respect to the origin or another fixed point.
  • A positive \( \phi \) shifts the graph to the left, while a negative \( \phi \) shifts it to the right.
• Real-world Implications: When modeling alternating currents, understanding the phase angle is crucial:
  • It indicates how waves such as voltage and current are synchronized, crucial for electrical engineering.
  • Phase differences can affect power transmission and signal processing.

In equations like the given exercise, the phase angle is essential for calculating root-mean-square voltage, helping determine the overall effectiveness of the alternating current system.

Integral Calculus

Integral calculus is the branch of calculus that deals with integrals, focusing on finding areas under curves and the accumulation of quantities. When engaging with problems like those involving root mean square voltage, integral calculus becomes invaluable.
Here's a deeper dive into the essentials:

• Basics of Integration:
  • Integration is often seen as the reverse process of differentiation. While differentiation breaks things down (to find slopes), integration builds up (to find areas).
  • The notation \( \int f(x) \, dx \) signifies the integral of function \( f(x) \) with respect to \( x \).
• Definite vs. Indefinite Integrals:
  • Indefinite Integrals: Provide the family of functions (antiderivatives) whose derivative yields the original function, plus a constant of integration \( C \).
  • Definite Integrals: Evaluate the area under a curve between two points, \( a \) and \( b \), denoted \( \int_a^b f(x) \, dx \). The result gives a numerical value representing the total accumulation within the limits.
• Use in RMS Voltage: In the exercise, integral calculus helps compute \( V_{\mathrm{rms}} \) by integrating the square of the sinusoidal voltage function over its period to find the average output.
  This concept is particularly useful in understanding how varying voltage affects heat production in electrical systems.

Mastering the fundamentals of integral calculus allows you to tackle complex mathematical problems efficiently, forming an essential tool in your mathematical toolbox.