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

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

The correct answer is B: 100 watts. The correct answer is B: 100 watts. A 200-volt peak-to-peak sine wave has a peak voltage of 100 volts and an RMS voltage of about 70.7 volts. Applying P = V(RMS)^2 / R to a 50-ohm load gives 70.7^2 / 50, or 100 watts PEP.

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Deep Dive: G5B06

The correct answer is B: 100 watts. A 200-volt peak-to-peak sine wave has a peak voltage of 100 volts and an RMS voltage of about 70.7 volts. Applying P = V(RMS)^2 / R to a 50-ohm load gives 70.7^2 / 50, or 100 watts PEP.

Why Other Answers Are Wrong

Option A: 1.4 watts results from an incorrect voltage conversion. Option C: 353.5 watts does not follow the RMS power calculation for a 50-ohm load. Option D: 400 watts incorrectly uses the peak-to-peak voltage directly in the power formula.

Exam Tip

Convert Vpp to Vpeak, then to Vrms: 200 / 2 / sqrt(2) = 70.7 V; square and divide by 50.

Memory Aid

200 Vpp into 50 ohms is 100 watts PEP.

Source & Coverage

Question Pool: 2023-2027 Question Pool

Subelement: G5B

Reference: 2023-2027 Question Pool · G5 - Electrical Principles

Key Concepts

peak envelope power peak-to-peak voltage RMS voltage 50-ohm load

Official Pool Source

Question text and answer key are synchronized with the current NCVEC General Class pool and mapped to the G5B topic.