Understanding Oscilloscope and Digital Multimeter Readings for Sine Waves

This question asks us to find the reading on a digital multimeter (DMM) when measuring an AC sine waveform that is observed on an oscilloscope. An oscilloscope displays the instantaneous voltage of a signal over time, showing its shape, frequency, and peak-to-peak voltage. A digital multimeter, when set to measure AC voltage, typically displays the RMS (Root Mean Square) value of the signal.

Analyzing the Given Sine Waveform Parameters

We are given the following information about the sine waveform:

• Frequency: 50 Hz (This information is not needed for the voltage calculation).
• Peak-to-peak voltage (\(V_{pp}\)): 20 V.

For a symmetrical sine wave, the peak voltage (\(V_p\)) is half of the peak-to-peak voltage.

So, the peak voltage is:

\begin{equation*} V_p = \frac{V_{pp}}{2} \end{equation*}

Substituting the given value:

\begin{equation*} V_p = \frac{20 \, \text{V}}{2} = 10 \, \text{V} \end{equation*}

Relating Oscilloscope Display to Digital Multimeter Reading

A standard digital multimeter measuring AC voltage on a sine wave usually displays the RMS voltage. The RMS voltage for a sine wave is calculated using the peak voltage:

\begin{equation*} V_{rms} = \frac{V_p}{\sqrt{2}} \end{equation*}

Using the peak voltage we found:

\begin{equation*} V_{rms} = \frac{10 \, \text{V}}{\sqrt{2}} \approx \frac{10}{1.414} \, \text{V} \approx 7.07 \, \text{V} \end{equation*}

Based on the standard understanding of DMMs measuring sine waves, the expected reading would be approximately 7.07 V.

Exploring the Provided Answer (6.36 V)

The provided correct answer is 6.36 V. This value is not the standard RMS voltage for a 10 V peak sine wave. However, it is related to the average value of a rectified sine wave. The average value of one half-cycle of a sine wave (or the average of a full-wave rectified sine wave) is calculated as:

\begin{equation*} V_{avg\_rectified} = \frac{2 V_p}{\pi} \end{equation*}

Let's calculate this value using the peak voltage \(V_p = 10 \, \text{V}\):

\begin{equation*} V_{avg\_rectified} = \frac{2 \times 10 \, \text{V}}{\pi} = \frac{20}{\pi} \, \text{V} \end{equation*}

Using the approximate value of \(\pi \approx 3.14159\):

\begin{equation*} V_{avg\_rectified} \approx \frac{20}{3.14159} \, \text{V} \approx 6.366 \, \text{V} \end{equation*}

This value, approximately 6.36 V, matches one of the options. While most modern digital multimeters are "true RMS" or "average-responding, RMS-calibrated" (which would read 7.07 V for a sine wave), some older or simpler meters might display a value closer to the average value, or the question might be based on an interpretation where the reading corresponds to this calculated average value.

Given the options and the provided correct answer, it appears the intended calculation leads to the average value of the rectified waveform.

Comparing with Options

Let's list the calculated values and compare them with the options:

• Peak-to-peak Voltage: 20 V
• Peak Voltage: 10 V
• Standard RMS Voltage (\(V_p/\sqrt{2}\)): ≈ 7.07 V
• Average of Rectified Voltage (\(2V_p/\pi\)): ≈ 6.36 V

The option that matches the calculated average value of the rectified waveform is 6.36 V.

Conclusion

Based on the provided options and the need to match the correct answer, the digital multimeter reading is considered to be 6.36 V, which corresponds to the average value of the rectified sine waveform.

Revision Table: Oscilloscope and Digital Multimeter

Additional Information: RMS vs. Average Voltage and DMM Types

When dealing with AC waveforms like a sine wave, different ways of measuring voltage exist. The most common are Peak, Peak-to-Peak, Average, and RMS.

• Peak Voltage (\(V_p\)): The maximum voltage reached from zero.
• Peak-to-Peak Voltage (\(V_{pp}\)): The difference between the maximum positive and maximum negative voltage levels. For a symmetric sine wave, \(V_{pp} = 2 V_p\).
• Average Voltage: The mathematical average of the voltage over time. For a complete cycle of a sine wave, the average voltage is zero. However, the term "average voltage" in the context of AC measurement often refers to the average of the absolute value of the voltage, typically over a half cycle or full cycle after rectification. For a sine wave, the average value of a half-cycle is \(\frac{2V_p}{\pi}\).
• RMS Voltage (Root Mean Square): This is the most common way to specify AC voltage because it relates the heating effect of the AC voltage to an equivalent DC voltage. For a sine wave, \(V_{rms} = \frac{V_p}{\sqrt{2}} \approx 0.707 V_p\).

Digital Multimeter Types:

• Average-Responding DMM: These meters measure the average value of the rectified AC waveform and then scale this reading to display the RMS value, assuming the waveform is a pure sine wave. They are accurate for sine waves but will give incorrect RMS readings for non-sinusoidal waveforms. The scaling factor is approximately 1.11 (\(\frac{\pi}{2\sqrt{2}}\)), which is the ratio of RMS to average for a sine wave. If such a meter were *not* scaled and simply displayed the average rectified value, it would read \(2V_p/\pi\), which is approximately 6.36V for \(V_p = 10V\).
• True RMS DMM: These meters use more complex circuitry to calculate the true RMS value of any AC waveform, whether it's sinusoidal or not. They provide accurate RMS readings for square waves, triangle waves, and other complex signals, as well as sine waves.

In typical scenarios involving standard digital multimeters and sine waves, the reading is the RMS value (7.07 V for a 10 V peak signal). However, the presence of 6.36 V as an option and the provided answer suggest a focus on the average rectified value calculation in this specific problem.