Problem 30 - Physics

Question

According to Equation \(20.7,\) an ac voltage \(V\) is given as a function of time \(t\) by \(V=V_{0} \sin 2 \pi f t,\) where \(V_{0}\) is the peak voltage and \(f\) is the frequency (in hertz). For a frequency of \(60.0 \mathrm{~Hz}\), what is the smallest value of the time at which the voltage equals one-half of the peak value?

Step-by-Step Solution

Answer

The smallest time is approximately 0.00139 seconds.

Step 1: Understand the Equation

The given equation for the ac voltage is \( V = V_0 \sin(2\pi f t) \). Here, \( V_0 \) represents the peak voltage, \( f \) is the frequency, and \( t \) is the time. We aim to find the smallest time \( t \) when voltage \( V \) is half of the peak value \( V_0 \).

Step 2: Set Up the Voltage Condition

We need \( V = \frac{1}{2}V_0 \). Substituting in the equation gives: \( \frac{1}{2} V_0 = V_0 \sin(2\pi f t) \).

Step 3: Solve for the Sine Function

Divide both sides by \( V_0 \) to isolate the sine function: \( \frac{1}{2} = \sin(2\pi f t) \).

Step 4: Find the Angle for the Sine

The sine function equals \( \frac{1}{2} \) at angles \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), but since we want the smallest positive time, we use \( \frac{\pi}{6} \).

Step 5: Calculate Time Using Frequency

We know \( 2\pi f t = \frac{\pi}{6} \). Solve for \( t \) using \( f = 60.0 \text{ Hz} \): \( 2\pi \times 60.0 \times t = \frac{\pi}{6} \).

Step 6: Complete the Time Calculation

Solve for \( t \): first, simplify \( 2 \times 60 \cdot t = \frac{1}{6} \) by dividing to get \( t = \frac{1}{720} \). Thus, \( t \approx 0.00139 \text{ s} \).

Key Concepts

Frequency

In the context of AC voltage, frequency refers to how often the electrical cycle repeats per second.
It is measured in hertz (Hz). For example, a frequency of 60 Hz means the voltage completes 60 cycles in one second.

Understanding frequency is crucial for time calculations in AC circuits.

• Higher frequency means more cycles in a given time.
• Lower frequency means fewer cycles, spreading the waveform over a longer period.

Frequency is directly tied to time calculations in problems involving AC voltage. It helps determine how quickly changes happen in the waveform.

Peak Voltage

Peak voltage, denoted as \( V_0 \), is the maximum voltage value reached during one cycle of an AC waveform.
It represents the highest point of the voltage swing in a sine wave.

This concept is important in understanding the amplitude of the voltage.

• Peak voltage is used in equations to describe the full extent of the voltage's potential difference.
• Understanding peak voltage helps you analyze when a specific voltage level is reached during the cycle.

In AC voltage calculations, knowing the peak voltage allows you to determine other points, like half of the peak voltage.

Sine Function

The sine function emerges from trigonometry and is critical in describing the shape of AC voltage.
In the equation \( V = V_0 \sin(2\pi ft) \), the sine function captures the oscillating nature of the voltage.

Key points on sine waves include:

• \( \sin(\theta) \) cycles between -1 and 1.
• Specific angle values, such as \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), give predictable results like \( \sin(\theta) = \frac{1}{2} \).

Identifying these angles helps in solving for time, where specific voltage levels are reached.

Time Calculation

To determine the time at which a certain voltage level is reached in an AC waveform, you employ time calculation.
You solve equations that involve frequency and a known voltage condition.

For example, solving \( 2\pi ft = \frac{\pi}{6} \), where frequency \( f = 60 \text{ Hz} \), involves isolating \( t \):

• Rearrange to find \( t = \frac{1}{720} \).
• This calculation reveals that the time is approximately 0.00139 seconds.

Time calculations are essential for predicting when certain voltage levels occur in AC circuits.