CRAC Class B (2025) — Questions 931–940
U1006. A resistive load has voltage U across it and consumes power P. Regarding the load's resistance R and the current I flowing through it, which is correct? ('x＾m' means 'x to the power m')
一个电阻负载两端电压为U，所消耗的功率为P。关于负载的电 阻R及流过其中的电流I，以下描述正确的是：（“x＾m”表示 “x的m次方”）
Exam tips · 考试提示
From P = U²/R we get R = U²/P, and from P = U·I we get I = P/U. Option A is correct; B uses R = U/P (wrong), and C/D invert.
U1007. Two loads have resistances R1 and R2. R1's resistance is N times R2's. After connecting them in parallel to a source, which is correct? ('x＾m' means 'x to the power m')
有阻值分别为R1和R2的两个负载。R1的阻值是R2的N倍。把 它们并联后接到电源上，则以下描述正确的是：（“x＾m”表 示“x的m次方”）
Exam tips · 考试提示
In parallel, both see the same voltage; I = U/R, so I1/I2 = R2/R1 = 1/N, and P = U²/R gives the same 1/N ratio. Option A is correct. B/C/D wrongly assume equal currents (that is the series case).
U1008. Two loads have resistances R1 and R2. R1's resistance is N times R2's. After connecting them in series to a source, which is correct? ('x＾m' means 'x to the power m')
有阻值分别为R1和R2的两个负载。R1的阻值是R2的N倍。把 它们串联后接到电源上，则以下描述正确的是：（“x＾m”表 示“x的m次方”）
Exam tips · 考试提示
In series the current is equal; P = I²R, so P1/P2 = R1/R2 = N. Option A is correct. B inverts the ratio, and C/D wrongly assume unequal currents.
U1009. Given that devices
已知A、B两个设备的工作电压相同，若A所消耗的电功率是B 的N倍，则以下描述正确的是：（“x＾m”表示“x的m次 方”）
Exam tips · 考试提示
With the same voltage, P = U·I, so current scales directly with power: I_A = N·I_B. Option A is correct. The N^(1/2) factor (B) applies only when comparing at fixed current/resistance, not here.
U1010. Given that devices
已知A、B两个设备的工作电压相同，若流过A的电流是B的N 倍，则以下描述正确的是：（“x＾m”表示“x的m次方”）
Exam tips · 考试提示
Same voltage, P = U·I, so if I_A = N·I_B then P_A = N·P_B. Option A is correct. The N² factor (C) would apply at fixed voltage comparing resistance (P = U²/R), not here.
U1011. After connecting N identical resistive loads in series to a source, compared with connecting each load separately to the source: ('x＾m' means 'x to the power m')
将N个相同的电阻负载串联后接到电源上，则与每个负载分别 接到电源上相比：（“x＾m”表示“x的m次方”）
Exam tips · 考试提示
Series total resistance is N·R, so current = U/(N·R) = (1/N) of the single case. Power P = I²R drops to (1/N)² = 1/N². Option A is correct.
U1012. After connecting N identical resistive loads in parallel to a source, compared with connecting each load separately to the source: ('x＾m' means 'x to the power m')
将N个相同的电阻负载并联后接到电源上，则与每个负载分别 接到电源上相比：（“x＾m”表示“x的m次方”）
Exam tips · 考试提示
In parallel each resistor still sees full source voltage, so its current and power are unchanged; with N of them the total power is N times a single one. Option A is correct.
U1013. For a resistive load, if the voltage across it is increased by n%, then: ('x＾m' means 'x to the power m')
对于一个电阻负载，若将其两端的电压提高n%，则：（“x＾ m”表示“x的m次方”）
Exam tips · 考试提示
P = U²/R, so raising U by n% gives U' = U(1+n/100) and P' = P(1+n/100)². Option A is correct; power scales with the square, not linearly.
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”）
Exam tips · 考试提示
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：
Exam tips · 考试提示
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.