Problem 8 - College Physics

Question

An ac voltage source has a peak voltage of 85 and a frequency of \(60 \mathrm{~Hz}\). The voltage at \(t=0\) is zero. (a) If a student measures the voltage at \(t=1 / 240 \mathrm{~s},\) how many possible results are there: (1) one, (2) two, or (3) three? Why? (b) Determine all possible voltages the student might measure.

Step-by-Step Solution

Answer

(a) One result, as the function is unique at the given time. (b) The voltage measured is 85 V.

Step 1: Understanding the Voltage Equation

The voltage of an AC source can be given by the equation \( V(t) = V_0 \sin(2\pi ft + \phi) \), where \( V_0 \) is the peak voltage, \( f \) is the frequency, and \( \phi \) is the phase constant. Since the voltage at \( t=0 \) is zero, \( \phi = 0 \), making the equation \( V(t) = 85 \sin(2\pi \times 60 \times t) \).

Step 2: Evaluate the Voltage at Given Time

Substitute \( t = \frac{1}{240} \mathrm{~s} \) into the equation. The voltage becomes: \[ V\left(\frac{1}{240}\right) = 85 \sin \left(2\pi \times 60 \times \frac{1}{240}\right) = 85 \sin \left(\frac{\pi}{2}\right) \]. Since \( \sin \left(\frac{\pi}{2}\right) = 1 \), this evaluates to \( 85 \).

Step 3: Determine the Periodicity of the Function

The AC voltage source is periodic with a period \( T = \frac{1}{f} = \frac{1}{60} \) seconds. At \( t = \frac{1}{240} \mathrm{~s} \), it is exactly a quarter of the period \( T \). Thus, any measurement made at \( t = \frac{1}{240} \mathrm{~s} + nT \) (where \( n \) is an integer) will also result in the same voltage.

Step 4: Analyze Potential Different Measurements

Given the periodicity of the sine function and the time point \( t \), the sine function is unique and results in a single measurement for this specific time due to its progression through one complete cycle over time. Thus, there's only one possible result in this duration.

Step 5: Conclude the Voltage Measurement

Since only one voltage measurement is possible at the given time when considering one complete cycle of the sine function, the voltage measured is 85 V.

Key Concepts

Peak Voltage

Understanding peak voltage is essential when studying AC Voltage. In an alternating current (AC) circuit, the voltage constantly changes direction and magnitude. Peak voltage, also known as maximum amplitude, is the highest value that AC voltage reaches during its cycle.

• Peak voltage is distinct from root mean square (RMS) voltage, which is a more typical measure in AC circuits because it represents the effective value that delivers the same power as DC voltage.
• In our problem, the peak voltage is given as 85 volts. This means the voltage will reach a maximum of 85 volts above and below zero during its cycle.
• The equation of the voltage function in terms of peak voltage is: \[ V(t) = V_0 \sin(2\pi f t + \phi) \]

In this formula, \( V_0 \) denotes the peak voltage, illustrating how far the AC voltage stretches from the midline (zero) to its highest and lowest points.

Frequency

Frequency in AC circuits defines how often the current changes direction or completes a full cycle per second. It is measured in hertz (Hz).

• For an AC circuit, a higher frequency means that the voltage completes more cycles per second and changes direction more rapidly.
• The problem specifies a frequency of 60 Hz, meaning the voltage completes 60 cycles every second.
• This frequency is common in household AC supply systems across many countries.

The frequency is vital to understanding how the voltage behaves over time, when plugged into the voltage equation \( V(t) = V_0 \sin(2\pi f t + \phi) \).
If the frequency is 60 Hz, then one complete cycle has a duration of \( \frac{1}{f} = \frac{1}{60} \) seconds, known as the period \( T \).
Knowing the frequency helps us predict voltage values at different times by analyzing how the sine wave progresses through its cycles.

Sine Wave Function

The sine wave function is a crucial concept when discussing alternating current (AC) voltages. This mathematical function describes the oscillating nature of AC voltage.

• The sine wave is characterized by its symmetric, smooth wave-like pattern.
• In the voltage equation \( V(t) = V_0 \sin(2\pi f t + \phi) \), the \( \sin \) term dictates how the voltage varies over time in a periodic manner.

For AC voltage, the sine wave reflects how the voltage continually rises and falls as it completes its cycle.

• The value \( \phi \) is the phase shift, determining where the wave begins at time \( t=0 \). In this scenario, it is zero, meaning the wave starts at zero voltage.

This wave function, complete with peaks and troughs, explains why measurements at certain times, such as \( t = \frac{1}{240} \), yield specific voltage results. For instance, at this time, the wave corresponds to a peak, making the voltage measure the peak value of 85 volts. Understanding these properties of the sine wave function helps clarify why certain values are possible at specific points in time.