CRAC Class C (2025) — Questions 991–1000
U1076. For an ideal transformer, the turns ratio of primary to secondary determines the voltage ratio. Suppose an ideal transformer has 1100 primary turns and 110 secondary turns, the primary is connected to a 220 V AC source, and a 22 Ω resistor is placed in parallel across the secondary. Which of the following conclusions is correct? (Voltages and currents are RMS values.) (Choose all that apply.)
理想变压器初、次级绕组的匝数比决定变压器的电压变比。假 设一个理想变压器的初级和次级绕组分别为1100匝和110匝， 将初级接入电压为220V的交流电源并在次级并联一个22欧的 电阻，则以下结论正确的是：（电压和电流均为有效值）
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Turns ratio = 1100:110 = 10:1, so V_s = 220/10 = 22 V (A). With 22 Ω load, I_s = 22/22 = 1 A (B). Reflected primary impedance = (N_p/N_s)²·R = 10²·22 = 2200 Ω = 2.2 kΩ (C). D is the trap: an ideal transformer's own 功耗 (power it dissipates) is ~0; the 22 W is the power delivered to the load, not the transformer's consumption.
U1077. For an ideal transformer, the turns ratio determines the voltage ratio. Suppose an ideal transformer has 120 primary turns and 1800 secondary turns, the primary is connected to a 220 V AC source, and a 4.5 kΩ resistor is placed in parallel across the secondary. Which of the following conclusions is correct? (Voltages and currents are RMS values.) (Choose all that apply.)
理想变压器初、次级绕组的匝数比决定变压器的电压变比。假 设一个理想变压器的初级和次级绕组分别为120匝和1800匝， 将初级接入电压为220V的交流电源并在次级并联一个4.5千欧 的电阻，则以下结论正确的是：（电压和电流均为有效值）
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Turns ratio = 120:1800 = 1:15, so V_s = 220·15 = 3300 V (A). Secondary current = 3300/4500 = 0.733 A; load power = 3300·0.733 ≈ 2420 W, equal to primary power, so I_p = 2420/220 = 11 A (B). Reflected primary impedance = (120/1800)²·4500 = (1/15)²·4500 = 4500/225 = 20 Ω (C). D is the trap: 2.42 kVA is the apparent power handled (load throughput), not the ideal transformer's own 功耗 (which is ~0).
U1078. In RF circuits, one use of a transformer is impedance transformation.
在射频电路中，变压器的用途之一是完成阻抗变换。某个绕制 于铁氧体磁环上的高频变压器，初、次级绕组的匝数分别为3匝 和6匝。在该变压器的设计工作频率和负载阻抗范围内，以下结 论正确的是：
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Turns ratio = 6/3 = 2. Impedance ratio = (turns ratio)² = 4, so it gives 1:4 or, by swapping windings, 4:1 (A). Voltage ratio = 2, giving 1:2 or 2:1 (B). Reversing a winding's leads inverts phase while keeping the ratio, acting as a phase inverter with transformation (C). D is false — a transformer cannot convert a reactive (imaginary) impedance component into a resistance.
U1079. If an antenna has an impedance of 450 Ω, what turns ratio transformer is needed to transform it to 50 Ω?
如果一副天线的阻抗为450欧，我们要用怎样匝数比的变压器 才能将之变换为50欧？
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Impedance ratio = 450/50 = 9, and since impedance scales as the square of the turns ratio, the turns ratio = √9 = 3:1 (A). A 9:1 turns ratio would give a 81:1 impedance ratio, so B is the common mistake of using the impedance ratio directly as the turns ratio.
U1080. A transmission-line transformer is an RF transformer designed using transmission-line theory and is widely used in amateur projects. For example, a few twisted-pair turns through a toroid make a 1:4 impedance transformer. The advantages of this kind of transformer are: (Choose all that apply.)
传输线变压器是结合传输线理论设计制作的一种射频变压器， 已广泛应用于业余无线电作品。例如，用双绞线在磁环上穿绕 数匝即可制成1:4阻抗变换器。这种变压器的优点是：
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Transmission-line transformers excel at wideband matching (A), are the basis of baluns (B), and can build power combiners/splitters (C). They perform impedance/frequency- ratio transformation, not frequency multiplication or division, so D is false.
U1081. Using a solid wire as a connection in an RF circuit at frequencies of tens of megahertz produces which phenomenon?
在频率为数十兆赫的射频电路中用实心导线作连线会有下列现 象：
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At tens of MHz, 趋肤效应 (skin effect) forces current into a thin surface layer of the conductor, so the interior carries almost no current (A). B describes DC behavior; C and D misdescribe the effect.
U1082. For solid copper wire and a hollow copper tube of the same material, diameter, and length, how do their heat losses in an AC circuit compare?
同样材料、同样直径、同样长度的实心铜线和空心铜管，在交 流电路中的发热损耗情况为：
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At low frequency, current uses the full cross-section, so the solid wire (more conductive material) has lower resistance/loss than the hollow tube. At high frequency, skin effect confines both to the surface; with the same outer diameter the surface area is the same, so their losses become equal (A).
U1083. A capacitor measures a capacitive reactance of 100 Ω at 50 Hz. One would infer its reactance at 5 MHz should be 0.001 Ω, but the measured value is significantly larger. This is because: (Choose all that apply.)
一个电容器，在50Hz频率下测得容抗为100欧姆。推断其在 5MHz下的容抗应为0.001欧姆，但实测值却显著偏大。这是因 为：
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The simple X_C = 1/(2πfC) ignoring parasitics predicts ~0.001 Ω at 5 MHz, but the capacitor's lead/electrode inductance adds series inductive reactance (A) and dielectric loss grows with frequency (B), both making the measured impedance larger. C/D (instrument flaw or formula breakdown) are not the real reasons.
U1084. An inductor coil measures an inductive reactance of 10 Ω at 50 Hz. One would infer its reactance at 5 MHz should be about 1 MΩ, but the measured value is significantly smaller. This is because:
一个电感线圈，在50Hz频率下测得感抗为10欧姆。推断其在 5MHz下的感抗应为1兆欧，但实测值却显著偏小。这是因为：
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The ideal X_L = 2πfL predicts ~1 MΩ at 5 MHz, but the coil's 匝 间分布电容 (inter-winding distributed capacitance) forms a parallel resonance that shunts the inductance, so the measured impedance is far smaller near/above self-resonance (A). B/C/D are not the cause.
U1085. Which of the following are active devices? (Choose all that apply.)
下列哪些属于有源器件？
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有源器件 (active devices) are components that can control current/voltage and provide gain or rectification via a non- linear transfer characteristic. Diodes (silicon rectifier, Schottky) and transistors (FET, BJT) all qualify, so A, B, C, D are correct.