Try to draw a sine wave and then mark off what peak-to-peak voltage is. Compute
the zero-to-peak voltage, 250V. Root mean square voltage is then 250*.707=176V.
176^2/50
On Mon, Apr 1, 2013 at 3:47 PM, citybadger <[email protected]> wrote:
> "What is the output PEP from a transmitter if an oscilloscope measures 500
> volts peak-to-peak across a 50-ohm resistor connected to the transmitter
> output?"
> The answer is 625 watts. How does one calculate that?
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