CRAC Class B (2025) — Questions 941–950
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：
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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.
U1024. For a square-wave signal with a minimum of 0 V and a peak- to-peak of 100 V, its average value is:
对于最小值为0伏，峰-峰值为100伏的方波信号，其平均值电 压为：
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A 0-to-100 V square wave (50% duty) has an average of 50.0 V (half the peak). Option A is correct. B is peak, C is the sine RMS factor.
U1025. For a triangular-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伏的三角波信号，其平均 值电压为：
Exam tips · 考试提示
A symmetric triangular wave (-50 V to +50 V) has equal positive and negative area, so its average over a full cycle is 0 V. Option A is correct.