CRAC Class C (2025) — Questions 891–900
Unofficial study reference only. Not affiliated with the FCC, NCVEC or CRAC — official sources prevail. 非官方学习参考，与 FCC、NCVEC、CRAC 无隶属关系；以官方发布为准。
U1014. For a resistive load, if the voltage across it is decreased by n%, then: ('x＾m' means 'x to the power m')
对于一个电阻负载，若将其两端的电压降低n%，则：（“x＾ m”表示“x的m次方”）
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By P = U²/R, lowering U by n% gives P' = P(1−n/100)². Option A is correct; the reduction is quadratic, not linear.
U1015. For a sinusoidal AC signal with a peak-to-peak value of 100 V, its RMS voltage is:
对于峰-峰值为100伏的正弦交流信号，其有效值电压为：
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Peak-to-peak 100 V → peak = 50 V → RMS = 50/√2 ≈ 35.4 V. Option A is correct. B is the RMS for a 100 V peak sine, C is the peak-to-peak of a 100 V peak, D is the half-amplitude.
U1016. For a sinusoidal AC signal with a peak value of 100 V, its RMS voltage is:
对于峰值为100伏的正弦交流信号，其有效值电压为：
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For a sine wave, RMS = peak/√2 = 100/1.414 ≈ 70.7 V. Option A is correct. B is the RMS of a 100 V peak-to-peak sine, C is the peak-to-peak.
U1017. For a square-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its RMS voltage is:
对于最小值为-50伏、峰-峰值为100伏的方波信号，其有效值 电压为：
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A symmetric square wave swinging -50 V to +50 V has an amplitude of 50 V; its RMS equals that amplitude, 50.0 V. Option A is correct. The 0.707 factor (B) belongs to sine waves, not square waves.
U1018. For a square-wave signal with a minimum of 0 V and a peak- to-peak of 100 V, its RMS voltage is:
对于最小值为0伏，峰-峰值为100伏的方波信号，其有效值电 压为：
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A 0-to-100 V square wave (50% duty) has an amplitude of 100 V and RMS = 50.0 V. Option A is correct. C is the sine factor, B is the peak.
U1019. For a triangular-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its RMS voltage is:
对于最小值为-50伏，峰-峰值为100伏的三角波信号，其有效 值电压为：
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The wave swings symmetrically from -50 V to +50 V (peak amplitude 50 V). For a triangular wave the RMS is lower than a sine's; with a 50 V peak the RMS is 25.0 V (i.e., peak/2). Option A matches. B is the square-wave value, C is the sine factor, D is a sine peak-to-peak value.
U1020. For a triangular-wave signal with a minimum of 0 V and a peak-to-peak of 100 V, its RMS voltage is:
对于最小值为0伏、峰-峰值为100伏的三角波信号，其有效值 电压为：
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The wave runs 0 to 100 V (peak amplitude 100 V). For a triangular wave the RMS is half the peak, i.e., 50.0 V. Option A matches. B is the peak, C is the sine factor.
U1021. For a sinusoidal AC signal with a peak-to-peak value of 100 V, its average value is:
对于峰-峰值为100伏的正弦交流信号，其平均值电压为：
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A full-cycle sine wave has equal positive and negative area, so its average (mean) over a full period is 0 V. Option A is correct. The other options are RMS or peak values, not the average.
U1022. For a sinusoidal AC signal with a peak value of 100 V, its average value is:
对于峰值为100伏的正弦交流信号，其平均值电压为：
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A full-cycle sine has average 0 V regardless of amplitude. Option A is correct; B/C/D are amplitude/RMS/peak-to-peak values.
U1023. For a square-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its average value is:
对于最小值为-50伏、峰-峰值为100伏的方波信号，其平均值 电压为：
Exam tips · 考试提示
A symmetric square wave (-50 V to +50 V, 50% duty) has equal positive and negative area, so its average is 0 V. Option A is correct.