What is the peak-to-peak RF voltage on the 50-ohm output of a 100 watt transmitter?

Study for the Federal Communications Commission (FCC) Element 3 Test. Use flashcards and multiple choice questions with hints and explanations. Get prepared for your exam today!

To determine the peak-to-peak RF voltage on a 50-ohm output of a 100-watt transmitter, it is important to understand the relationship between power, voltage, and resistance. The key formula to use here relates power (P), voltage (V), and resistance (R):

[ P = \frac{V_{\text{rms}}^2}{R} ]

Where:

  • ( P ) is the power in watts (100 watts in this case),

  • ( V_{\text{rms}} ) is the root mean square voltage,

  • ( R ) is the resistance in ohms (50 ohms here).

From this formula, we can isolate ( V_{\text{rms}} ):

[ V_{\text{rms}} = \sqrt{P \times R} ]

Substituting in the values:

[ V_{\text{rms}} = \sqrt{100 \times 50} = \sqrt{5000} \approx 70.71 , \text{volts} ]

The peak voltage (( V_{\text{peak}} )) is related to the RMS voltage by the relationship:

[ V_{\text{peak}} = V_{\

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