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

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

• A. 100 volts
• B. 150 volts
• C. 200 volts
• D. 250 volts

Answer: (C)

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_{\

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_{\