FCC Exam Question: 3-10B1

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

A. 70 volts. C. 140 volts.
B. 100 volts. D. 200 volts.
C.
D.
Correct Answer: D

Explanation: To determine the peak-to-peak RF voltage, we first calculate the RMS voltage using the power (P) and resistance (R). The formula for RMS voltage (V_RMS) is: V_RMS = sqrt(P * R) V_RMS = sqrt(100 W * 50 ohms) = sqrt(5000) = 70.71 volts For a sinusoidal waveform, the peak voltage (V_peak) is V_RMS multiplied by the square root of 2 (approximately 1.414): V_peak = V_RMS * sqrt(2) = 70.71 V * 1.414 = 100 volts Finally, the peak-to-peak voltage (V_p-p) is twice the peak voltage: V_p-p = V_peak * 2 = 100 V * 2 = 200 volts Therefore, the peak-to-peak RF voltage is approximately 200 volts. Option A (70 volts) is close to the RMS voltage. Option B (100 volts) is the peak voltage. Option C (140 volts) is approximately twice the RMS voltage, but not the peak-to-peak.

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Includes Elements 1, 3, 6, 7R, 8, and 9.