Peak Diode Voltmeter and Direct Voltage

A peak diode voltmeter is designed to measure the peak value of an input signal. When a voltage is applied to this type of voltmeter, it charges a capacitor to the peak voltage of the signal. The meter then reads this peak voltage.

However, the scale of this particular voltmeter is calibrated to display the Root Mean Square (RMS) voltage of a sine wave. For a pure sine wave, the relationship between the RMS value ($V_{rms}$) and the peak value ($V_{peak}$) is given by:

\(V_{rms} = \frac{V_{peak}}{\sqrt{2}}\)

This means the voltmeter internally calculates and displays \(V_{peak} / \sqrt{2}\), even though it measured \(V_{peak}\).

In this problem, a direct voltage (DC voltage) is applied to the meter. A direct voltage has a constant value, let's call it \(V_{DC}\). The peak value of a direct voltage is simply the voltage itself, so \(V_{peak} = V_{DC}\).

The meter reads 36 Vrms. Since the meter is calibrated for a sine wave, the reading displayed (36 V) corresponds to the measured peak voltage divided by \(\sqrt{2}\).

So, we can write the equation based on the meter's calibration:

\(\text{Meter Reading} = \frac{V_{peak}}{\sqrt{2}}\)

We know the Meter Reading is 36 V and, for a direct voltage input, \(V_{peak} = V_{DC}\). Substituting these values into the equation:

\(36 \, \text{V} = \frac{V_{DC}}{\sqrt{2}}\)

Now, we need to solve for \(V_{DC}\):

\(V_{DC} = 36 \, \text{V} \times \sqrt{2}\)

Using the approximate value \(\sqrt{2} \approx 1.414\):

\(V_{DC} \approx 36 \times 1.414\)

\(V_{DC} \approx 50.904 \, \text{V}\)

Rounding this value to the nearest integer gives 51 V.

Therefore, the value of the applied direct voltage is approximately 51 V.

Let's summarize the steps:

1. Identify that the voltmeter measures the peak voltage.
2. Recognize that the input is a direct voltage, so its peak value equals its DC value (\(V_{peak} = V_{DC}\)).
3. Understand that the meter scale is calibrated to show \(V_{peak}/\sqrt{2}\) for a sine wave.
4. Set the given meter reading (36 V) equal to the calibrated output formula: \(36 = V_{peak}/\sqrt{2}\).
5. Substitute \(V_{peak} = V_{DC}\) into the equation: \(36 = V_{DC}/\sqrt{2}\).
6. Solve for \(V_{DC}\): \(V_{DC} = 36 \times \sqrt{2}\).
7. Calculate the value: \(V_{DC} \approx 36 \times 1.414 \approx 50.904\) V.

The closest option to this calculated value is 51 V.