Problem 63 - Contemporary Precalculus

Question

In an alternating current circuit, the voltage is given by the formula $$V=V_{\max } \cdot \sin (2 \pi f t+\phi)$$ where \(V_{\max }\) is the maximum voltage, \(f\) is the frequency (in cycles per second), \(t\) is the time in seconds, and \(\phi\) is the phase angle. (a) If the phase angle is \(0,\) solve the voltage equation for \(t\) (b) If \(\phi=0, V_{\max }=20, V=8.5,\) and \(f=120,\) find the smallest positive value of \(t\)

Step-by-Step Solution

Answer

Answer: The smallest positive value for time (t) is approximately \(5.81 \times 10^{-4} s\).

Step 1: (a) Solve the voltage equation for t when ϕ=0

When the phase angle ϕ is 0, the voltage equation becomes: $$V=V_{\max }\cdot \sin(2\pi ft)$$ Now, we can rearrange the equation to solve for t. Divide both sides of the equation by the maximum voltage \(V_{\max}\): $$\frac{V}{V_{\max }}=\sin(2\pi ft)$$ Next, take the inverse sine (arcsin) of both sides of the equation: $$t = \frac{\arcsin(\frac{V}{V_{\max }})}{2\pi f}$$ Now we have the equation for time (t) when the phase angle ϕ is 0.

Step 2: (b) Find the smallest positive value for t with specified parameter values

Given the parameter values \(\phi=0, V_{\max }=20, V=8.5,\) and \(f=120,\) we can substitute them into the equation for time (t) that we derived in part (a). $$t = \frac{\arcsin(\frac{8.5}{20})}{2\pi(120)}$$ Now we manually calculate the value of this equation. First, we calculate the fraction inside the arcsin function. $$\frac{8.5}{20} = 0.425$$ Then, we take the inverse sine of this value. $$\arcsin(0.425)\approx 0.4393$$ Finally, we divide this result by the \(2\pi f\) value. $$t \approx \frac{0.4393}{2\pi(120)} \approx 5.81 \times 10^{-4} s$$ However, as we know, the sine function has multiple solutions for one value. In this case, we have a second solution as well. Remember that the sine function has a period of \(2\pi\) and: $$\sin(x) = \sin(\pi - x)$$ So, we can find the second solution for t as follows: $$t_{2} = \frac{\pi - 0.4393}{2\pi(120)} \approx 2.175 \times 10^{-3} s$$ As the smallest positive value of t is required, we can conclude that t is approximately: $$t \approx 5.81 \times 10^{-4} s$$

Key Concepts

Voltage Equation

The concept of the voltage equation in an alternating current (AC) circuit is central to understanding how voltage varies over time. In an AC circuit, the voltage is not constant; it oscillates in a sinusoidal manner. The voltage at any point in time can be calculated using the equation: \[ V = V_{\max} \cdot \sin(2 \pi f t + \phi) \] Here:

• \( V \) is the instantaneous voltage
• \( V_{\max} \) is the maximum voltage, which is the peak value of the waveform
• \( f \) is the frequency, which represents how many times per second the waveform repeats
• \( t \) is time, measured in seconds
• \( \phi \) is the phase angle, indicating the waveform's horizontal displacement

This equation illustrates how the voltage output changes, highlighting the roles of different parameters.

Trigonometric Equations

Trigonometric equations are equations involving trigonometric functions, such as sine, cosine, and tangent. They are crucial in solving problems related to AC circuits, like finding the time at specific voltage values. When dealing with the voltage equation in AC circuits: \[ \sin(2 \pi f t + \phi) = \frac{V}{V_{\max}} \] We typically need to solve it for time \( t \). This involves using the inverse trigonometric function, 'arcsin.' The process is:

• Reorganize the equation to isolate the trigonometric component
• Use the arcsine function to "undo" the sine, which allows solving for the angle inside the sine function
• Translate this angle into a timeframe, considering the frequency \( f \)

Trigonometric functions can yield multiple solutions due to their periodic nature, so care is needed to explore all suitable time values.

Phase Angle

The phase angle \( \phi \) in an AC circuit is a measure of displacement between the waveform at a certain point in time and a reference point.

• If the phase angle is zero, it implies that at \( t = 0 \), the waveform starts in a standard position without any shift.
• A positive phase angle means the waveform shifts to the left.
• A negative phase angle indicates a shift to the right.

In applications, the phase angle can result from circuit elements like resistors, capacitors, and inductors, affecting how current and voltage relate over time. This concept helps in comprehending why real-world signals might not perfectly align, despite having the same frequency.

Frequency in AC Circuits

Frequency \( f \) in AC circuits represents the number of complete cycles that occur in one second. It is measured in hertz (Hz) and defines how quickly the waveform oscillates. For example, a frequency of 60 Hz indicates the waveform goes through 60 cycles per second.

• Frequency influences how fast changes occur in the circuit.
• It is crucial for determining how often the voltage completes a cycle, affecting appliances' operation that depend on AC power.
• In an equation like \( \sin(2 \pi f t + \phi) \), frequency influences the period, or the time it takes for one full wave cycle. This is the reciprocal of the frequency: \( \text{Period} = \frac{1}{f} \).

Understanding frequency helps predict and synchronize the performance of various electronic components in a practical context.