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LaTeX is fine just use \$ around it.
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edit approved Feb 15, 2015 at 7:09
Dan D.
edit approved Feb 15, 2015 at 7:09
Dan D.
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Taking an educated guess here, I'd start from the origin of this V_{RMS} = V_{pp} \frac{1}{\sqrt{2}} (sorry for the LaTeX notation here)\$V_{RMS} = V_{pp} \frac{1}{\sqrt{2}}\$

This formula holds only for the sine wave - it's actually the formula for the voltage that would deliver the same average equivalent power to the resistive load. It can be derived from a reasonably simple integration of the instantanous power.

Putting any non-resistive component (such as a diode) in the circuit, causes that integral to no longer hold. Specifically, around 0, the diode would isolate entirely (in both directions), and thus there would be no power contribution to the resistive load. For a simple model of a diode, you could probably work out the exact formula yourself with a little bit of integration.

Taking an educated guess here, I'd start from the origin of this V_{RMS} = V_{pp} \frac{1}{\sqrt{2}} (sorry for the LaTeX notation here)

This formula holds only for the sine wave - it's actually the formula for the voltage that would deliver the same average equivalent power to the resistive load. It can be derived from a reasonably simple integration of the instantanous power.

Putting any non-resistive component (such as a diode) in the circuit, causes that integral to no longer hold. Specifically, around 0, the diode would isolate entirely (in both directions), and thus there would be no power contribution to the resistive load. For a simple model of a diode, you could probably work out the exact formula yourself with a little bit of integration.

Taking an educated guess here, I'd start from the origin of this \$V_{RMS} = V_{pp} \frac{1}{\sqrt{2}}\$

This formula holds only for the sine wave - it's actually the formula for the voltage that would deliver the same average equivalent power to the resistive load. It can be derived from a reasonably simple integration of the instantanous power.

Putting any non-resistive component (such as a diode) in the circuit, causes that integral to no longer hold. Specifically, around 0, the diode would isolate entirely (in both directions), and thus there would be no power contribution to the resistive load. For a simple model of a diode, you could probably work out the exact formula yourself with a little bit of integration.

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qdot
qdot
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Taking an educated guess here, I'd start from the origin of this V_{RMS} = V_{pp} \frac{1}{\sqrt{2}} (sorry for the LaTeX notation here)

This formula holds only for the sine wave - it's actually the formula for the voltage that would deliver the same average equivalent power to the resistive load. It can be derived from a reasonably simple integration of the instantanous power.

Putting any non-resistive component (such as a diode) in the circuit, causes that integral to no longer hold. Specifically, around 0, the diode would isolate entirely (in both directions), and thus there would be no power contribution to the resistive load. For a simple model of a diode, you could probably work out the exact formula yourself with a little bit of integration.