Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

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: Calculate Average Voltage over 1 Second

The average voltage over 1 second is the integral of the voltage function over 1 second divided by the time interval (1 second).\[\text{Average voltage} = \frac{1}{1 - 0} \int_0^1 \hat{V} \sin(120\pi t + \phi)\, dt\]Since the function is a pure sine wave, the average over a full period is 0.Hence, the average voltage is 0 volts.

Step 2: Calculate Average Voltage over 1/60 Second

The average voltage over \(\frac{1}{60}\) second is the integral of the voltage function over this smaller interval divided by the interval.\[\text{Average voltage} = \frac{1}{1/60} \int_0^{1/60} \hat{V} \sin(120\pi t + \phi)\, dt\]This evaluates to zero as the integral of a full sine wave cycle over any complete cycle or its fraction is 0.Thus, the average voltage over \(1/60\) second remains 0 volts.

Step 3: Compute the Integral for RMS Voltage

We use the formula for \(V_{\mathrm{rms}}\), which involves integrating the square of the function over 1 second.\[V_{\mathrm{rms}} = \sqrt{\frac{1}{1} \int_{0}^{1} (\hat{V} \sin(120 \pi t + \phi))^2 \, dt}\]Using the hint, we integrate \(\int \sin^2(x) \, dx = \frac{1}{2}t - \frac{1}{4}\sin(2x) + C\). Thus:\[\int_0^{1} \sin^2(120 \pi t + \phi) \, dt = \frac{1}{2} \]Substituting back:\[V_{\mathrm{rms}} = \hat{V} \sqrt{\frac{1}{2}}\]Thus, \( V_{\mathrm{rms}} = \frac{\hat{V} \sqrt{2}}{2} \).

Step 4: Solve for Maximum Voltage \(\hat{V}\)

Given \( V_{\mathrm{rms}} = 120 \) volts:\[120 = \frac{\hat{V} \sqrt{2}}{2}\]Solving for \( \hat{V} \):\[\hat{V} = 120 \times \frac{2}{\sqrt{2}} = 120\sqrt{2} \approx 169.7 \text{ volts}\]Thus, the maximum voltage \(\hat{V}\) is approximately 169.7 volts.

Key Concepts

Voltage Oscillation

In alternating current (AC) systems, the voltage oscillates regularly over time. This oscillation is typically represented by a sine wave, such as the function given by \( V = \hat{V} \sin(120 \pi t + \phi) \).
This function describes how the voltage changes with time \( t \), where:

• \( \hat{V} \) is the peak voltage, the maximum value the voltage can reach at any point in time.
• \( 120\pi \) reflects the frequency of the oscillation; it means the voltage completes 60 cycles per second, a typical frequency in household electricity systems.
• \( \phi \) is the phase angle, which adjusts the position of the sine curve along the time axis.

Voltage oscillation is essential in AC systems because it allows efficient energy transfer over large distances, which is why electric companies use AC voltage. The changing voltage induces current flow, powering electrical devices. Understanding the oscillatory behavior of voltage helps in comprehending various real-life applications, from the lights in our homes to complex industrial machines.

Root-Mean-Square Voltage

The root-mean-square (RMS) voltage is a crucial concept in understanding AC systems. It gives us a way to quantify the effective voltage in an AC circuit, analogous to the direct current (DC) equivalent. The formula for RMS voltage \( V_{\mathrm{rms}} \) is:
\[ V_{\mathrm{rms}} = \sqrt{\int_{\phi}^{1+\phi}(\hat{V} \sin(120 \pi t+\phi))^{2} dt} \]
The significance of RMS voltage lies in its ability to represent the effective power potential of the AC wave, as it calculates the square root of the average of the squares of instantaneous voltages over a complete cycle.

• The RMS value is crucial for determining how much work, i.e., energy, can be done by the AC voltage.
• It provides a standard way to compare with DC systems, facilitating engineers in building effective systems.

In household terms, the RMS voltage would typically be noted as 120 volts, indicative of its capacity to perform work comparable to a 120-volt DC circuit. With the derived equation \( V_{\mathrm{rms}} = \frac{\hat{V} \sqrt{2}}{2} \), one can relate the maximum potential in terms of peak voltage \( \hat{V} \). This formula reveals the inherent efficiency of AC systems and ensures the correct delivery of electric energy in a manageable form.

Integral Calculus

Integral Calculus is a branch of calculus that deals with integrals and their properties. In the context of the given exercise, integral calculus is utilized to solve problems involving continuous waveforms, such as the voltage oscillations described.
The calculation of RMS voltage specifically requires integration:

• You measure the total area under the squared voltage curve \((\hat{V} \sin(120 \pi t + \phi))^2\) over the interval \([\phi, 1+\phi]\).
• This involves evaluating the integral \( \int \sin^2(x) \ dx \), which is solved using techniques and rules from integral calculus.
• Applying these calculations, specifically approximations and the periodicity of sine functions, allows simplifying the expression, as shown in \( V_{\mathrm{rms}} = \hat{V} \sqrt{\frac{1}{2}} \).

Integral calculus helps not just in technical calculations on paper but also in understanding the nature of oscillations and signals in practical settings, making it vital for fields that range from engineering to applied physics. Comprehending basic integral principles is pivotal for students and professionals working on anything involving dynamic systems.