FCC Exam Question: 6A529
By what factor must the voltage of an ac circuit, as indicated on the scale of an AC voltmeter, bemultiplied to obtain the average voltage value?
Explanation: An AC voltmeter typically measures and displays the Root Mean Square (RMS) voltage for a sinusoidal waveform. This RMS value represents the "effective" voltage that would produce the same amount of heat in a resistive load as a comparable DC voltage. The "average voltage value" for a symmetrical AC waveform like a sine wave, calculated over one complete cycle, is **zero**. This is because the positive half-cycle exactly cancels out the negative half-cycle when averaged over the full period. Therefore, to obtain the average voltage value (which is 0) from an AC voltmeter reading (which is typically the RMS value), you must multiply the RMS reading by a factor of **0**. * **A) 0:** This is correct. The average voltage of a full sine wave cycle is 0. * **B) 0:** This is a duplicate of A, also correct. * **C) By the factor of** and **D) By the factor of**: These options are incomplete but imply a non-zero factor. If the question had asked for the *average rectified voltage* (average over a half-cycle or of the absolute value), the factor would be approximately 0.9 (since V_avg_half-cycle ≈ 0.637 * V_peak and V_peak ≈ 1.414 * V_rms, so V_avg_half-cycle ≈ 0.9 * V_rms). However, the question simply asks for the "average voltage value," which, by standard definition for a full cycle of a sine wave, is zero.
6A299
6A63
6A315
6A134
6A68
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Includes Elements 1, 3, 6, 7R, 8, and 9.