Ham Amateur Radio Technician Practice Exam

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Prepare for the Ham Amateur Radio Technician Exam. Study using flashcards and multiple choice questions with detailed explanations. Boost your confidence and readiness today!

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What is the PEP produced by 200 volts peak-to-peak across a 50-ohm dummy load?

50 watts
75 watts
100 watts
150 watts

The correct answer is: 100 watts

To determine the Peak Envelope Power (PEP) produced by a voltage of 200 volts peak-to-peak across a 50-ohm dummy load, we first need to convert the peak-to-peak voltage into effective peak voltage. The effective peak voltage (V_peak) can be calculated as follows: V_peak = V_peak-to-peak / 2 = 200 volts / 2 = 100 volts Now, we can use the formula for power in a resistive load, which is given by: Power (P) = V_peak^2 / R Where: - P is the power in watts, - V_peak is the voltage across the load in volts, - R is the resistance of the load in ohms. Substituting the values we have: P = (100 volts)^2 / 50 ohms P = 10000 volts^2 / 50 ohms P = 200 watts However, since we want to find the average power, considering both positive and negative half cycles in the context of PEP in an amplitude modulation scenario where the power is effectively doubled due to the modulation process, we arrive at the following: PEP = 200 watts / 2 = 100