CRAC Bilingual Manual › Part: Electrical Basics for Amateur Radio
CRAC Bilingual Exam Manual (Class A / B / C) | 中国业余无线电台操作技术能力验证英中对照手册
This section covers Ohm's Law and Resistance with 67 bilingual questions from the CRAC 2025 question bank. Each question shows the original Chinese (left) and the English translation (right). The correct answer is highlighted in green, followed by a Knowledge Point Analysis and Candidate Tips covering US–China differences, common pitfalls, and real on-air practice.
Class badges ABC indicate which license-class syllabus includes each question. Class A is the entry level, Class B adds HF privileges, and Class C is the advanced level.
将一个电阻为R的负载接到电压为U的电源上。关于负载中的电流I及负载所消耗的功率Ｐ，以下描述正确的是：（“x＾m”表示“x的m次方”）
- AI=U/R；P=U＾2/R
- BI=U/R；P=U/R
- CI=R/U；P=U＾2×R
- DI=R/U；P=U×R
Connect a load of resistance R to a source of voltage U. Regarding the current I in the load and the power P consumed by the load, which of the following descriptions is correct? (‘x＾m’ means ‘x to the power m’)
- AI = U/R; P = U^2/R
- BI = U/R; P = U/R
- CI = R/U; P = U^2 × R
- DI = R/U; P = U × R
By Ohm’s law I = U/R, and power P = U·I = U²/R. Option A gives both correctly. B wrongly uses P = U/R, and C/D swap the resistance relationship.
US–China difference: Same Ohm’s law in the US.
Common pitfall: Confusing P = U²/R with P = U/R.
Real on-air practice: Compute heatsink needs from P = U²/R.
一个电阻为R的负载中流过的电流为I。关于负载两端的电压U及负载所消耗的功率Ｐ，以下描述正确的是：（“x＾m”表示“x的m次方”）
- AU=I×R；P=I＾2×R
- BU=I×R；P=I×R
- CU=R / I；P=R / I＾2
- DU=R / I；P=R / I
A current I flows through a load of resistance R. Regarding the voltage U across the load and the power P consumed by the load, which of the following is correct? (‘x＾m’ means ‘x to the power m’)
- AU = I × R; P = I^2 × R
- BU = I × R; P = I × R
- CU = R / I; P = R / I^2
- DU = R / I; P = R / I
U = I·R (Ohm’s law) and P = U·I = I²·R. Option A is correct; B misstates power, and C/D invert the relations.
US–China difference: Same.
Common pitfall: Writing P = I·R instead of I²·R.
Real on-air practice: A 2 A current through 50 Ω dissipates 200 W — match your resistor rating.
一个电阻负载两端电压为U，流过的电流为I。关于该负载的电阻R和所消耗的功率P，以下描述正确的是：（“x＾m”表示“x的m次方”）
- AR=U/I；P=U×I
- BR=U×I；P=U/I
- CR=U×I；P=U/ I＾2
- DR=U×I；P=U/ I
A resistive load has voltage U across it and current I flowing through it. Regarding the load’s resistance R and consumed power P, which is correct? (‘x＾m’ means ‘x to the power m’)
- AR = U/I; P = U × I
- BR = U × I; P = U/I
- CR = U × I; P = U / I^2
- DR = U × I; P = U / I
R = U/I and P = U·I are the basic definitions, so A is correct. Options B/C/D wrongly equate R with U·I.
US–China difference: Same.
Common pitfall: Mixing up P = U·I with R = U·I.
Real on-air practice: At 13.8 V drawing 10 A, the power is 138 W.
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”）
- AR=U＾2/P；I=P/U
- BR=U/P；I=P/U
- CR=P＾2/U；I=U/P
- DR=P/U；I=U/P
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’)
- AR = U^2/P; I = P/U
- BR = U/P; I = P/U
- CR = P^2/U; I = U/P
- DR = P/U; I = U/P
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.
US–China difference: Same.
Common pitfall: Swapping U/P for U²/P.
Real on-air practice: A 50 Ω dummy load at 100 W has U = √(50·100) ≈ 70.7 V.
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”）
- A流过R1的电流是R2的1/N，R1消耗的功率是R2的1/N
- B流过R1的电流是R2的N倍，R1消耗的功率是R2的N＾2倍
- C流过R1的电流与R2的相同，R1消耗的功率是R2的1/N＾2
- D流过R1的电流与R2的相同，R1消耗的功率是R2的N倍
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’)
- Athe current through R1 is 1/N of R2’s, and the power consumed by R1 is 1/N of R2’s
- Bthe current through R1 is N times R2’s, and the power consumed by R1 is N^2 times R2’s
- Cthe current through R1 is the same as R2’s, and the power consumed by R1 is 1/N^2 of R2’s
- Dthe current through R1 is the same as R2’s, and the power consumed by R1 is N times R2’s
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).
US–China difference: Same.
Common pitfall: Thinking parallel currents are equal (that is series).
Real on-air practice: Two parallel 50 Ω loads each draw an equal current share.
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”）
- A流过R1的电流与R2的相同，R1消耗的功率是R2的N倍
- B流过R1的电流与R2的相同，R1消耗的功率是R2的1/N
- C流过R1的电流是R2的1/N，R1消耗的功率是R2的1/N＾2
- D流过R1的电流是R2的N倍，R1消耗的功率是R2的N＾2倍
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’)
- Athe current through R1 is the same as R2’s, and R1 consumes N times the power of R2
- Bthe current through R1 is the same as R2’s, and R1 consumes 1/N the power of R2
- Cthe current through R1 is 1/N of R2’s, and R1 consumes 1/N^2 the power of R2
- Dthe current through R1 is N times R2’s, and R1 consumes N^2 the power of R2
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.
US–China difference: Same.
Common pitfall: Thinking series currents differ.
Real on-air practice: In a series string the bigger resistor dissipates more heat.
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”）
- AA的工作电流是B的N倍
- BA的工作电流是B的N＾(1/2)倍
- CA的工作电流是B的N＾2倍
- DA的工作电流是B的1/N倍
Given that devices A and B operate at the same voltage, if A consumes N times the electrical power of B, which is correct? (‘x＾m’ means ‘x to the power m’)
- AA’s operating current is N times B’s
- BA’s operating current is N^(1/2) times B’s
- CA’s operating current is N^2 times B’s
- DA’s operating current is 1/N times B’s
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.
US–China difference: Same.
Common pitfall: Using √N (that applies if current is fixed and you compare voltage).
Real on-air practice: A 100 W amp at 13.8 V draws 10× the current of a 10 W one.
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”）
- AA所消耗的电功率是B的N倍
- BA所消耗的电功率是B的N＾(1/2)倍
- CA所消耗的电功率是B的N＾2倍
- DA所消耗的电功率是B的1/N倍
Given that devices A and B operate at the same voltage, if the current through A is N times that through B, which is correct? (‘x＾m’ means ‘x to the power m’)
- Athe power consumed by A is N times B’s
- Bthe power consumed by A is N^(1/2) times B’s
- Cthe power consumed by A is N^2 times B’s
- Dthe power consumed by A is 1/N times B’s
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.
US–China difference: Same.
Common pitfall: Squaring when you should not (N² only if comparing resistance at fixed voltage).
Real on-air practice: Doubling current at fixed voltage doubles power.
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”）
- A串联后流过每个电阻的电流减少到1/N，每个电阻的耗电功率减少到1/N＾2
- B串联后流过每个电阻的电流减少到1/N，每个电阻的耗电功率减少到1/N
- C串联后流过每个电阻的电流不变，每个电阻的耗电功率减少到1/N
- D串联后流过每个电阻的电流增加到N倍，每个电阻的耗电功率增加到N＾2倍
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’)
- Ain series, the current through each resistor drops to 1/N, and each resistor’s power consumption drops to 1/N^2
- Bin series, the current through each resistor drops to 1/N, and each resistor’s power consumption drops to 1/N
- Cin series, the current through each resistor is unchanged, and each resistor’s power consumption drops to 1/N
- Din series, the current through each resistor increases to N times, and each resistor’s power consumption increases to N^2 times
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.
US–China difference: Same.
Common pitfall: Forgetting current drops to 1/N, so power drops to 1/N².
Real on-air practice: Series lamps are each dimmer than one alone.
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”）
- A并联后流过每个电阻的电流不变，所有电阻的总耗电功率为单个电阻的N倍
- B并联后流过每个电阻的电流不变，所有电阻的总耗电功率为单个电阻的N＾2倍
- C并联后流过每个电阻的电流增加到N倍，每个电阻的总耗电功率增加到N＾2倍
- D并联后流过每个电阻的电流减少到1/N，每个电阻的总耗电功率减少到1/N＾2
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’)
- Ain parallel, the current through each resistor is unchanged, and the total power consumption of all resistors is N times that of a single resistor
- Bin parallel, the current through each resistor is unchanged, and the total power consumption is N^2 times that of a single resistor
- Cin parallel, the current through each resistor increases to N times, and each resistor’s power increases to N^2 times
- Din parallel, the current through each resistor drops to 1/N, and each resistor’s power drops to 1/N^2
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.
US–China difference: Same.
Common pitfall: Thinking per-resistor current changes in parallel.
Real on-air practice: Adding parallel loads increases the total current drawn from the supply.
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”）
- A耗电量增加到原来的[（100+n）/100]＾2
- B耗电量增加到原来的（100+n）/100
- C耗电量比原来增加n%
- D耗电量比原来增加（n%）＾2
For a resistive load, if the voltage across it is increased by n%, then: (‘x＾m’ means ‘x to the power m’)
- Apower consumption increases to [(100+n)/100]^2 of the original
- Bpower consumption increases to (100+n)/100 of the original
- Cpower consumption increases by n% over the original
- Dpower consumption increases by (n%)^2 over the original
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.
US–China difference: Same.
Common pitfall: Thinking power scales linearly with voltage (it is squared).
Real on-air practice: A 10% voltage rise means about 21% more heat.
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”）
- A耗电量减少到原来的[（100-n）/100]＾2
- B耗电量减少到原来的（100-n）/100
- C耗电量比原来减少n%
- D耗电量比原来减少（n%）＾2
For a resistive load, if the voltage across it is decreased by n%, then: (‘x＾m’ means ‘x to the power m’)
- Apower consumption decreases to [(100-n)/100]^2 of the original
- Bpower consumption decreases to (100-n)/100 of the original
- Cpower consumption decreases by n% from the original
- Dpower consumption decreases by (n%)^2 from the original
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.
US–China difference: Same.
Common pitfall: Linear assumption.
Real on-air practice: A brownout reduces heating quadratically.
对于峰-峰值为100伏的正弦交流信号，其有效值电压为：
- A35.4伏
- B70.7伏
- C141伏
- D50.0伏
For a sinusoidal AC signal with a peak-to-peak value of 100 V, its RMS voltage is:
- A35.4 V
- B70.7 V
- C141 V
- D50.0 V
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.
US–China difference: Same.
Common pitfall: Using the peak (70.7) instead of peak-to-peak/2 then /√2.
Real on-air practice: Know the 0.707 and 0.354 factors for sine waves.
对于峰值为100伏的正弦交流信号，其有效值电压为：
- A70.7伏
- B35.4伏
- C141伏
- D50.0伏
For a sinusoidal AC signal with a peak value of 100 V, its RMS voltage is:
- A70.7 V
- B35.4 V
- C141 V
- D50.0 V
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.
US–China difference: Same.
Common pitfall: Using peak-to-peak (141) or half (50).
Real on-air practice: A 100 V peak sine reads 70.7 V RMS on the meter.
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：
- A50.0伏
- B70.7伏
- C35.4伏
- D100伏
For a square-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its RMS voltage is:
- A50.0 V
- B70.7 V
- C35.4 V
- D100 V
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.
US–China difference: Same.
Common pitfall: Applying the sine RMS factor (0.707) to a square wave.
Real on-air practice: A 50 V square keying tone has 50 V RMS.
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：
- A50.0伏
- B100伏
- C70.7伏
- D35.4伏
For a square-wave signal with a minimum of 0 V and a peak-to-peak of 100 V, its RMS voltage is:
- A50.0 V
- B100 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same.
Common pitfall: Using 70.7 for a square wave.
Real on-air practice: A 0–100 V 50% square is 50 V RMS.
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：
- A25.0伏
- B50.0伏
- C70.7伏
- D35.4伏
For a triangular-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its RMS voltage is:
- A25.0 V
- B50.0 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same waveform relationships are taught in the US.
Common pitfall: Applying the sine 0.707 factor to a triangle wave.
Real on-air practice: Triangle/sawtooth test tones run cooler than sine for the same peak.
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：
- A50.0伏
- B100伏
- C70.7伏
- D35.4伏
For a triangular-wave signal with a minimum of 0 V and a peak-to-peak of 100 V, its RMS voltage is:
- A50.0 V
- B100 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same.
Common pitfall: Using the sine 0.707 factor on a triangle.
Real on-air practice: A 0–100 V triangle ramp is 50 V RMS.
对于峰-峰值为100伏的正弦交流信号，其平均值电压为：
- A0伏
- B35.4伏
- C70.7伏
- D141伏
For a sinusoidal AC signal with a peak-to-peak value of 100 V, its average value is:
- A0 V
- B35.4 V
- C70.7 V
- D141 V
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.
US–China difference: Same.
Common pitfall: Confusing average with RMS.
Real on-air practice: DC-blocking coupling means the average is 0.
对于峰值为100伏的正弦交流信号，其平均值电压为：
- A0伏
- B50.0伏
- C70.7伏
- D35.4伏
For a sinusoidal AC signal with a peak value of 100 V, its average value is:
- A0 V
- B50.0 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same.
Common pitfall: Giving the RMS (70.7) as the average.
Real on-air practice: The average of any symmetric AC is 0 over a full cycle.
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：
- A0伏
- B50.0伏
- C70.7伏
- D50.0伏
For a square-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its average value is:
- A0 V
- B50.0 V
- C70.7 V
- D50.0 V
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.
US–China difference: Same.
Common pitfall: Giving the RMS (50 V) as the average.
Real on-air practice: A symmetric square keying tone averages to 0 V DC.
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：
- A50.0伏
- B100伏
- C70.7伏
- D35.4伏
For a square-wave signal with a minimum of 0 V and a peak-to-peak of 100 V, its average value is:
- A50.0 V
- B100 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same.
Common pitfall: Giving RMS (50 V happens to equal average here, but the reasoning differs for other waves).
Real on-air practice: A 0–100 V 50% square has a 50 V DC average.
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：
- A0伏
- B50.0伏
- C70.7伏
- D35.4伏
For a triangular-wave signal with a minimum of -50 V and a peak-to-peak of 100 V, its average value is:
- A0 V
- B50.0 V
- C70.7 V
- D35.4 V
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.
US–China difference: Same.
Common pitfall: Confusing average with RMS for a triangle.
Real on-air practice: A symmetric triangle has zero DC average.
对于最小值为0伏、峰-峰值为100伏的三角波信号，其平均值电压为：
- A50.0伏
- B100伏
- C70.7伏
- D35.4伏
For a triangular-wave signal with a minimum of 0 V and a peak-to-peak of 100 V, its average value is:
- A50.0 V
- B100 V
- C70.7 V
- D35.4 V
A 0-to-100 V triangle (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.
US–China difference: Same.
Common pitfall: Mixing up average and RMS for a triangle.
Real on-air practice: A 0–100 V triangle ramp averages 50 V.
用电压为120V的蓄电池组和峰值电压为120V的交流变压器分别驱动参数相同的两个电阻负载。在相同时间内，哪一个电阻发出的热量多？
- A蓄电池驱动的电阻所发的热量是交流变压器上的电阻的2倍左右
- B蓄电池驱动的电阻所发的热量是交流变压器上的电阻的0.7倍左右
- C蓄电池驱动的电阻所发的热量是交流变压器上的电阻的1.4倍左右
- D两个电源所驱动的电阻发热相同
Drive two identical resistive loads with a 120 V battery bank and an AC transformer of 120 V peak voltage respectively. Over the same time, which resistor emits more heat?
- Athe heat from the battery-driven resistor is about 2 times that of the resistor on the AC transformer
- Bthe heat from the battery-driven resistor is about 0.7 times that of the resistor on the AC transformer
- Cthe heat from the battery-driven resistor is about 1.4 times that of the resistor on the AC transformer
- Dthe two resistors emit the same heat
The battery supplies 120 V DC → power = 120²/R. The AC transformer’s peak is 120 V, so its RMS = 120/√2 ≈ 84.9 V → power = 84.9²/R = 7200/R. The ratio 14400/7200 = 2, so the battery-driven resistor emits about twice the heat. Option A is correct.
US–China difference: Same.
Common pitfall: Comparing AC peak to DC directly instead of using RMS.
Real on-air practice: A DC supply at the same number as an AC peak delivers more heat.
用电压为120V的蓄电池组和有效值电压为120V的交流变压器分别驱动参数相同的两个电阻负载。在相同时间内，哪一个电阻发出的热量多？
- A两个电源所驱动的电阻发热相同
- B蓄电池驱动的电阻所发的热量是交流变压器上的电阻的1.4倍左右
- C蓄电池驱动的电阻所发的热量是交流变压器上的电阻的0.7倍左右
- D蓄电池驱动的电阻所发的热量是交流变压器上的电阻的2倍左右
Drive two identical resistive loads with a 120 V battery bank and an AC transformer of 120 V RMS voltage respectively. Over the same time, which resistor emits more heat?
- Athe two resistors emit the same heat
- Bthe heat from the battery-driven resistor is about 1.4 times that of the resistor on the AC transformer
- Cthe heat from the battery-driven resistor is about 0.7 times that of the resistor on the AC transformer
- Dthe heat from the battery-driven resistor is about 2 times that of the resistor on the AC transformer
Both sources are 120 V RMS equivalent (the battery’s 120 V DC and the transformer’s 120 V RMS give the same effective voltage), so each resistor dissipates the same power and emits the same heat. Option A is correct.
US–China difference: Same.
Common pitfall: Thinking DC and “120 V” AC differ in heating (they don’t, at equal RMS).
Real on-air practice: A 120 V DC heater and a 120 V RMS AC heater get equally hot.
用电压为120V的蓄电池组和有效值电压为120V的交流变压器串联二极管后分别驱动参数相同的两个电阻负载。在相同时间内，哪一个电阻发出的热量多？（忽略二极管的正向压降）
- A蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的2倍左右
- B蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的1.4倍左右
- C蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的0.7倍左右
- D两个电源所驱动的电阻发热相同
Drive two identical resistive loads with a 120 V battery bank and a 120 V RMS AC transformer with a series diode respectively (ignore the diode forward drop). Over the same time, which resistor emits more heat?
- Athe heat from the battery-driven resistor is about 2 times that of the resistor in the AC transformer circuit
- Bthe heat from the battery-driven resistor is about 1.4 times that of the resistor in the AC transformer circuit
- Cthe heat from the battery-driven resistor is about 0.7 times that of the resistor in the AC transformer circuit
- Dthe two resistors emit the same heat
The battery gives 120 V DC. The AC (120 V RMS, peak ≈ 169.7 V) through a series diode is half-wave rectified; the load sees only the positive halves, with RMS = Vpk/2 ≈ 84.9 V. Power ratio = (120²)/(84.9²) = 2, so the battery-driven resistor is about twice as hot. Option A is correct.
US–China difference: Same.
Common pitfall: Ignoring that a series diode halves the effective voltage.
Real on-air practice: A diode in series with a lamp cuts its brightness roughly in half (heat ~1/4).
用电压为120V的蓄电池组和峰值电压为120V的交流变压器经过带电容滤波的全波整流电路分别驱动参数相同的两个电阻负载。在相同时间内，哪一个电阻发出的热量多？（忽略整流器的正向压降）
- A两个电源所驱动的电阻发热大致相同
- B蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的2倍左右
- C蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的1.4倍左右
- D蓄电池驱动的电阻所发的热量是交流变压器电路上的电阻的0.7倍左右
Drive two identical resistive loads with a 120 V battery bank and a 120 V peak AC transformer through a capacitor-filtered full-wave rectifier respectively (ignore the rectifier forward drop). Over the same time, which resistor emits more heat?
- Athe heat from the two is roughly the same
- Bthe heat from the battery-driven resistor is about 2 times that of the resistor in the AC transformer circuit
- Cthe heat from the battery-driven resistor is about 1.4 times that of the resistor in the AC transformer circuit
- Dthe heat from the battery-driven resistor is about 0.7 times that of the resistor in the AC transformer circuit
A capacitor-filtered full-wave rectifier charges to near the AC peak (≈120 V, ignoring drop), which equals the 120 V battery DC. Thus both resistors see ~120 V and dissipate roughly the same power. Option A is correct.
US–China difference: Same.
Common pitfall: Thinking rectified DC equals RMS (it is ~peak).
Real on-air practice: A cap-filtered supply charges to near the AC peak.
用有效值电压为120V、频率为50Hz的交流电源和有效值电压为120V、频率为10kHz的方波电源分别驱动参数相同的两个电阻负载。在相同时间内，哪一个电阻发出的热量多？
- A两个电源所驱动的电阻发热大致相同
- B10kHz电路电阻所发的热量是50Hz电路电阻的5倍左右
- C10kHz电路电阻所发的热量是50Hz电路电阻的1/5左右
- D10kHz电路电阻所发的热量是50Hz电路电阻的200倍左右
Drive two identical resistive loads with a 120 V RMS, 50 Hz AC source and a 120 V RMS, 10 kHz square-wave source respectively. Over the same time, which resistor emits more heat?
- Athe two resistors emit roughly the same heat
- Bthe 10 kHz resistor emits about 5 times the heat of the 50 Hz resistor
- Cthe 10 kHz resistor emits about 1/5 the heat of the 50 Hz resistor
- Dthe 10 kHz resistor emits about 200 times the heat of the 50 Hz resistor
A pure resistor’s heating depends only on the RMS voltage, not on frequency. Both sources are 120 V RMS, so both resistors dissipate the same power and emit the same heat. Option A is correct.
US–China difference: Same.
Common pitfall: Thinking a higher frequency means more heat in a resistor.
Real on-air practice: A 10 kHz square and a 50 Hz sine at the same RMS heat a resistor equally.
能够以电场形式储存能量的元件是：
- A电容
- B电阻
- C压敏元件
- D电感
The component capable of storing energy in the form of an electric field is:
- Aa capacitor
- Ba resistor
- Ca varistor
- Dan inductor
A 电容 (capacitor) stores energy in an electric field between its plates. A resistor dissipates energy as heat; an inductor stores energy in a magnetic field; a varistor is a voltage-dependent nonlinear resistor. Only the capacitor stores energy electrostatically.
US–China difference: Same basic component physics in the US; capacitor/inductor energy storage is a universal concept.
Common pitfall: Confusing electric-field storage (capacitor) with magnetic-field storage (inductor).
Real on-air practice: Your tuner and mic capacitor store charge; handle high-voltage caps with care after power-off.
若在电容器的两端施加一个正弦交流电压，则流过电容器的电流的大小：
- A与电压和电容量都成正比
- B与电压和电容量都成反比
- C与电压成正比，与电容量成反比
- D与电容量成正比，与电压成反比
If a sinusoidal AC voltage is applied across a capacitor, the magnitude of the current flowing through the capacitor:
- Ais directly proportional to both the voltage and the capacitance
- Bis inversely proportional to both the voltage and the capacitance
- Cis directly proportional to the voltage and inversely proportional to the capacitance
- Dis directly proportional to the capacitance and inversely proportional to the voltage
For a capacitor the current is I = V·ω·C = V·2πf·C, where V is the applied voltage and C is the capacitance. Thus the current magnitude is directly proportional to both the voltage and the capacitance, so A is correct and B/C/D are wrong.
US–China difference: Same formula I = 2πfCV used in US amateur theory.
Common pitfall: Forgetting the frequency factor and thinking current is inversely proportional to capacitance (as in the DC charging-time case).
Real on-air practice: A larger coupling capacitor passes more RF current at a given frequency.
一个电容器，在某一频率下测得容抗为若干欧姆。如果频率提高N倍，其容抗将：
- A减少到原来的1/N
- B增大到原来的N倍
- C减少到原来的1/（2πN）
- D增大到原来的2πN倍
For a capacitor whose capacitive reactance is measured as some number of ohms at a certain frequency, if the frequency is increased N-fold, its capacitive reactance will:
- Adecrease to 1/N of the original
- Bincrease to N times the original
- Cdecrease to 1/(2πN) of the original
- Dincrease to 2πN times the original
容抗 (capacitive reactance) is Xc = 1/(2πfC). It is inversely proportional to frequency, so raising f by a factor N reduces Xc to 1/N of its original value. A is correct; B reverses the relation, and C/D wrongly insert 2π.
US–China difference: Identical Xc = 1/(2πfC) relation used in US exams.
Common pitfall: Mixing up capacitive reactance (falls with frequency) with inductive reactance (rises with frequency).
Real on-air practice: At VHF/UHF a small capacitor looks nearly like a short to RF.
能够以磁场形式储存能量的元件是：
- A电感
- B电阻
- C电磁铁
- D电容
The component capable of storing energy in the form of a magnetic field is:
- Aan inductor
- Ba resistor
- Can electromagnet
- Da capacitor
An 电感 (inductor) stores energy in a magnetic field built up by current through its coil. A resistor dissipates energy; a capacitor stores electric-field energy; an electromagnet is a specialized coil but the generic energy-storage component named here is the inductor.
US–China difference: Same concept; US exams pair “magnetic field storage = inductor” with “electric field storage = capacitor”.
Common pitfall: Confusing magnetic-field (inductor) with electric-field (capacitor) storage.
Real on-air practice: Your loading coil stores magnetic energy while tuning a short antenna.
若在线圈两端施加一个正弦交流电压，则流过线圈的电流的大小：
- A与电压成正比，与电感量成反比
- B与电压和电感量都成正比
- C与电压和电感量都成反比
- D与电感量成正比，与电压成反比
If a sinusoidal AC voltage is applied across a coil, the magnitude of the current flowing through the coil:
- Ais directly proportional to the voltage and inversely proportional to the inductance
- Bis directly proportional to both the voltage and the inductance
- Cis inversely proportional to both the voltage and the inductance
- Dis directly proportional to the inductance and inversely proportional to the voltage
For an inductor I = V/(ωL) = V/(2πfL). The current magnitude is directly proportional to the applied voltage and inversely proportional to the inductance L, so A is correct.
US–China difference: Same relation I = V/(2πfL) in US material.
Common pitfall: Thinking larger inductance gives more current — it gives less (more reactance).
Real on-air practice: A bigger choke passes less RF current at a given frequency.
一个电感线圈，在某一频率下测得感抗为若干欧姆。如果频率提高N倍，其感抗将：
- A增大到原来的N倍
- B减少到原来的1/N
- C增大到原来的2πN倍
- D减少到原来的1/（2πN）
For an inductive coil whose inductive reactance is measured as some ohms at a certain frequency, if the frequency is increased N-fold, its inductive reactance will:
- Aincrease to N times the original
- Bdecrease to 1/N of the original
- Cincrease to 2πN times the original
- Ddecrease to 1/(2πN) of the original
感抗 (inductive reactance) is XL = 2πfL, directly proportional to frequency. Raising f by N makes XL N times larger. A is correct; B is the capacitive behavior, and C/D wrongly insert 2π.
US–China difference: Same XL = 2πfL in US exams.
Common pitfall: Swapping the direction vs capacitive reactance (which falls with frequency).
Real on-air practice: A 10 mH choke is a much bigger RF block at 430 MHz than at 14 MHz.
术语“相位差”通常用来描述频率相同的两个周期信号在时间上的超前或滞后关系。关于周期信号的相位差，以下描述正确的是：
- A两个信号的相位差为180°的偶数倍时为同相关系，为奇数倍时为反相关系
- B相位相差90°的两个信号互为正交关系
- C信号通过含有电感或电容的电路，其电压与电流将不再同相
- D无线电波因相互叠加而相互干涉，同相相长，反相相消
The term “phase difference” is typically used to describe the lead or lag relationship in time between two periodic signals of the same frequency. Which of the following descriptions of the phase difference of periodic signals is correct? (Choose all that apply.)
- AWhen the phase difference is an even multiple of 180° the signals are in phase; an odd multiple means they are in anti-phase
- BTwo signals differing in phase by 90° are in quadrature (orthogonal) with each other
- CWhen a signal passes through a circuit containing inductance or capacitance, its voltage and current are no longer in phase
- DRadio waves interfere by superposition: in-phase components reinforce, anti-phase components cancel
相位差 (phase difference): 0/360°/even multiples of 180° = in-phase (同相); odd multiples of 180° = anti-phase (反相) (A). A 90° difference is quadrature (正交) (B). Reactive elements (L/C) make voltage and current out of phase (C). Superposition of same-frequency waves gives constructive (in-phase) or destructive (anti-phase) interference (D). All four are correct.
US–China difference: Same phase vocabulary (in-phase, quadrature, anti-phase) in US material.
Common pitfall: Forgetting that 360° phase difference equals 0° (in phase), not anti-phase.
Real on-air practice: Phasing two antennas 90° apart builds a directional pattern (quadrature phasing).
将两个幅度相等，相位相差360°的正弦电压源串联，所得结果是：
- A幅度为单个信号源的2倍、相位与原信号源相同的正弦电压
- B电压为0
- C幅度与单个信号源的相同、相位与原信号源相差180°的正弦电压
- D幅度与单个信号源的相同、频率比原信号高一倍的正弦电压
Connecting two sinusoidal voltage sources of equal amplitude and with a phase difference of 360° in series yields:
- Aa sinusoidal voltage with amplitude twice that of a single source and the same phase as the original source
- Bzero voltage
- Ca sinusoidal voltage with the same amplitude as a single source and 180° different in phase from the original
- Da sinusoidal voltage with the same amplitude as a single source and double the original frequency
A 360° phase difference is equivalent to 0° — the two equal sources are in phase, so series addition doubles the amplitude while keeping the same phase. A is correct; B is the 180° (anti-phase) result, C/D are wrong.
US–China difference: Same phasor addition principle taught in US electronics.
Common pitfall: Treating 360° as if it were 180° and getting zero.
Real on-air practice: Stacking two identical in-phase drivers increases level, not cancel it.
将两个幅度相等，相位相差180°的正弦电压源串联，所得结果是：
- A电压为0
- B幅度为单个信号源的2倍、相位与原信号源相同的正弦电压
- C幅度与单个信号源的相同、相位与原信号源相差90°的正弦电压
- D幅度与单个信号源的相同、频率比原信号高一倍的正弦电压
Connecting two sinusoidal voltage sources of equal amplitude and with a phase difference of 180° in series yields:
- Azero voltage
- Ba sinusoidal voltage with amplitude twice that of a single source and the same phase as the original
- Ca sinusoidal voltage with the same amplitude as a single source and 90° different in phase from the original
- Da sinusoidal voltage with the same amplitude as a single source and double the original frequency
Two equal, 180°-apart (anti-phase) sources in series cancel completely, giving 0 V (A). This is the destructive-interference case; B is the in-phase result.
US–China difference: Same cancellation principle in US theory.
Common pitfall: Adding magnitudes instead of accounting for the 180° sign.
Real on-air practice: Two identical signals arriving equal and opposite cancel (e.g., noise-canceling phased systems).
将两个幅度相等，相位相差90°的正弦电压源串联，所得结果是：
- A幅度为单个信号源的1.41倍、相位与原信号源各差45°的正弦电压
- B幅度与单个信号源的相同、相位与原信号源相差45°的正弦电压
- C幅度为单个信号源的2倍、相位与原信号源相同的正弦电压
- D幅度与单个信号源的相同、频率比原信号高一倍的正弦电压
Connecting two sinusoidal voltage sources of equal amplitude and with a phase difference of 90° in series yields:
- Aa sinusoidal voltage with amplitude 1.41 times that of a single source and 45° different in phase from each original source
- Ba sinusoidal voltage with the same amplitude as a single source and 45° different in phase from the original
- Ca sinusoidal voltage with amplitude twice that of a single source and the same phase as the original
- Da sinusoidal voltage with the same amplitude as a single source and double the original frequency
Two equal sources 90° apart add by the Pythagorean theorem: resultant amplitude = √(V²+V²) = 1.41 V (√2). The result lies exactly halfway in phase, i.e. 45° from each. A is correct.
US–China difference: Same vector (phasor) addition in US courses.
Common pitfall: Guessing 2× (in-phase) or 0 (anti-phase) instead of √2.
Real on-air practice: A 90° hybrid combiner produces a √2 increase in combined level.
在电容器两端施加正弦交流电压，则流过其中的电流是正弦交流电流。该电流的相位：
- A超前于电压相位90度
- B落后于电压相位90度
- C与电压相位相同
- D与电压相差180度
When a sinusoidal AC voltage is applied across a capacitor, the current through it is a sinusoidal AC current. The phase of this current:
- Aleads the voltage phase by 90°
- Blags the voltage phase by 90°
- Cis the same as the voltage phase
- Ddiffers from the voltage by 180°
In a capacitor the current leads the voltage by 90° (电流超前电压90°). This follows from I = C·dV/dt. A is correct; an inductor is the opposite (current lags).
US–China difference: Same “ELI the ICE man” mnemonic (I leads E in a Capacitor) in the US.
Common pitfall: Reversing lead/lag for capacitor vs inductor.
Real on-air practice: In a dipole feed the current/voltage phase matters for matching.
若正弦交流电流流过电容器，则其两端的电压是正弦交流电压。该电压的相位：
- A落后于电流相位90度
- B超前于电流相位90度
- C与电压相位相同
- D与电流相差180度
If a sinusoidal AC current flows through a capacitor, the voltage across it is a sinusoidal AC voltage. The phase of this voltage:
- Alags the current phase by 90°
- Bleads the current phase by 90°
- Cis the same as the voltage phase
- Ddiffers from the current by 180°
This is the same capacitor relationship viewed from the other side: since current leads voltage by 90°, the voltage lags the current by 90°. A is correct and consistent with U1165.
US–China difference: Same concept; just restated as voltage-lags-current.
Common pitfall: Picking “leads” — that contradicts current-leading-voltage.
Real on-air practice: In a PI-network output the cap voltage phase shifts the load.
在线圈两端施加正弦交流电压，则流过其中的电流是正弦交流电流。该电流的相位：
- A落后于电压相位90度
- B超前于电压相位90度
- C与电压相位相同
- D与电压相差180度
When a sinusoidal AC voltage is applied across a coil, the current through it is a sinusoidal AC current. The phase of this current:
- Alags the voltage phase by 90°
- Bleads the voltage phase by 90°
- Cis the same as the voltage phase
- Ddiffers from the voltage by 180°
In an inductor (coil) the current lags the voltage by 90° (电流滞后电压90°), because the induced EMF opposes changes in current. A is correct; the capacitor is the opposite.
US–China difference: Same “ELI the ICE man” (E leads I in an Inductor) in the US.
Common pitfall: Assigning the capacitor’s lead relationship to the inductor.
Real on-air practice: A transmitter’s tank coil current lags its applied voltage.
若正弦交流电流流过线圈，则其两端的电压是正弦交流电压。该电压的相位：
- A超前于电流相位90度
- B落后于电流相位90度
- C与电流相位相同
- D与电流相差180度
If a sinusoidal AC current flows through a coil, the voltage across it is a sinusoidal AC voltage. The phase of this voltage:
- Aleads the current phase by 90°
- Blags the current phase by 90°
- Cis the same as the current phase
- Ddiffers from the current by 180°
Inductor restated from the current’s viewpoint: since current lags voltage by 90°, the voltage leads the current by 90°. A is correct and consistent with U1167.
US–China difference: Same concept, just restated.
Common pitfall: Picking “lags” — that would contradict current-lagging-voltage.
Real on-air practice: The induced voltage in a tuning coil leads its current.
将电阻R和电容C串联后突然接到直流电压U上，电容C两端的电压会：
- A从0按指数规律逐渐增加到U
- B从U按指数规律逐渐减小到0
- C从0突然跳到U，然后再按指数规律逐渐减小到0
- D从U突然跳到0，然后再按指数规律逐渐增大到U
An R–C series circuit is suddenly connected to a DC voltage U; the voltage across capacitor C will:
- Arise from 0 to U exponentially (following an exponential law)
- Bfall from U to 0 exponentially
- Cjump from 0 to U instantly, then fall to 0 exponentially
- Djump from U to 0 instantly, then rise to U exponentially
On suddenly connecting an RC series to DC, the capacitor voltage starts at 0 (it cannot change instantly) and charges toward U along an exponential curve Vc = U(1−e^(−t/τ)). A is correct; discharging (B) is the disconnect case.
US–China difference: Same RC charging equation taught in US courses.
Common pitfall: Thinking the capacitor voltage jumps instantly — only current/voltage across the resistor jump.
Real on-air practice: Keying a DC line into a decoupling cap charges it gradually.
给定电阻R和电容C，则其阻值和电容量的乘积称为时间常数τ。将R和C串联后突然接到直流电压U上，电容C两端的电压经过τ时间后大约为U的：
- A63%
- B99%
- C37%
- D6.28%
Given a resistor R and capacitor C, the product of the resistance and capacitance is called the time constant τ. When R and C are connected in series and suddenly connected to DC voltage U, after time τ the voltage across capacitor C is approximately what fraction of U?
- A63%
- B99%
- C37%
- D6.28%
After one time constant τ = RC, the capacitor reaches 1 − e^(−1) ≈ 0.632 = 63% of the final voltage U. A is correct; 37% is the remaining gap (the value for the decaying quantity at τ).
US–China difference: Same 63%/37% rule (1−e⁻¹, e⁻¹) in US material.
Common pitfall: Picking 37% (the leftover) instead of 63% (the reached value).
Real on-air practice: A keying envelope’s rise time is measured in τ units.
将电阻R和电容C串联后突然接到直流电压U上，电阻R两端的电压会：
- A从0突然跳到U，然后再按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到U
- C从U按直线规律逐渐减小到0
- D从U突然跳到0，然后再按直线规律逐渐减小到U
When an R–C series circuit is suddenly connected to a DC voltage U, the voltage across resistor R will:
- Ajump from 0 to U instantly, then fall to 0 exponentially
- Brise from 0 to U exponentially
- Cfall from U to 0 linearly
- Djump from U to 0 instantly, then decrease to U linearly
At the instant of connection the capacitor is uncharged (acts like a short), so the full U appears across R; then as the capacitor charges, VR = U·e^(−t/τ) decays exponentially to 0. A is correct.
US–China difference: Same RC transient behavior in US theory.
Common pitfall: Assigning the capacitor’s rising curve to the resistor.
Real on-air practice: The resistor sees a momentary full-voltage spike at switch-on.
给定电阻R和电容C，则其阻值和电容量的乘积称为时间常数τ。将R和C串联后突然接到直流电压U上，电阻R两端的电压经过τ时间后大约为U的：
- A37%
- B99%
- C63%
- D6.28%
Given a resistor R and capacitor C, the product RC is the time constant τ. When connected in series and suddenly connected to DC voltage U, after time τ the voltage across resistor R is approximately what fraction of U?
- A37%
- B99%
- C63%
- D6.28%
After one τ the resistor voltage has decayed to e^(−1) ≈ 0.368 ≈ 37% of its initial U. A is correct — this is the complement of the capacitor’s 63% at the same instant.
US–China difference: Same 37% decay value in US courses.
Common pitfall: Confusing 37% (resistor decay) with 63% (capacitor rise).
Real on-air practice: Switch-on surge across R drops to ~37% after one τ.
将电阻R和电容C串联后突然接到直流电压U上，流过电阻R的电流会：
- A从0突然跳到U/R，然后再按指数规律逐渐减小到0
- B从0突然跳到U/R并保持
- C从0按指数规律逐渐增加到U/R
- D从U/R突然跳到0并保持
When an R–C series circuit is suddenly connected to DC voltage U, the current through resistor R will:
- Ajump from 0 to U/R instantly, then fall to 0 exponentially
- Bjump from 0 to U/R instantly and stay there
- Crise from 0 to U/R exponentially
- Djump from U/R to 0 instantly and stay there
Current equals resistor voltage divided by R, so it jumps to U/R at t=0 (capacitor short) and then decays as I = (U/R)·e^(−t/τ) to 0 as the capacitor charges. A is correct. (The branch current through R and C is the same series current.)
US–China difference: Same RC current transient in US theory.
Common pitfall: Thinking current stays at U/R (that would be a steady DC, not charging).
Real on-air practice: The in-rush current into a filter cap is largest at switch-on.
将电阻R和电容C串联后突然接到直流电压U上，流过电容C的电流会：
- A从0突然跳到U/R，然后再按指数规律逐渐减小到0
- B从0突然跳到U/R并保持
- C从0按指数规律逐渐增加到U/R
- D从U/R突然跳到0并保持
When an R–C series circuit is suddenly connected to DC voltage U, the current through capacitor C will:
- Ajump from 0 to U/R instantly, then fall to 0 exponentially
- Bjump from 0 to U/R instantly and stay there
- Crise from 0 to U/R exponentially
- Djump from U/R to 0 instantly and stay there
In a series circuit the capacitor current equals the resistor current; it jumps to U/R at t=0 and decays to 0 as the capacitor finishes charging (I = C·dVc/dt → 0 when Vc stops changing). A is correct.
US–China difference: Same series-current identity in US courses.
Common pitfall: Thinking a DC capacitor current persists — it falls to zero at steady state.
Real on-air practice: A blocking cap passes only the transient/AC current, not steady DC.
将电阻R和电容C并联后接到电压为U的直流电源上。突然断开电源，电容C两端的电压会：
- A从U按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到U
- C从0突然跳到U并保持
- D从U突然跳到0并保持
An R–C parallel circuit is connected to a DC supply of voltage U. When the supply is suddenly disconnected, the voltage across capacitor C will:
- Afall from U to 0 exponentially
- Brise from 0 to U exponentially
- Cjump from 0 to U instantly and stay there
- Djump from U to 0 instantly and stay there
With the supply removed, the charged capacitor discharges through the parallel resistor: Vc = U·e^(−t/τ), decaying exponentially from U to 0. A is correct; the capacitor voltage cannot jump (it is stored energy).
US–China difference: Same parallel RC discharge in US material.
Common pitfall: Thinking the capacitor instantly loses its charge on disconnect.
Real on-air practice: A filtering cap bleeds down through its load/resistor after power-off.
将电阻R和电容C并联后接在电压为U的直流电源上。突然断开电源，电阻R两端的电压会：
- A从U按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到U
- C从0突然跳到U并保持
- D从U突然跳到0并保持
An R–C parallel circuit is connected to a DC supply of voltage U. When the supply is suddenly disconnected, the voltage across resistor R will:
- Afall from U to 0 exponentially
- Brise from 0 to U exponentially
- Cjump from 0 to U instantly and stay there
- Djump from U to 0 instantly and stay there
In parallel, the resistor and capacitor share the same voltage, so on disconnect the resistor voltage also decays exponentially from U to 0 along with the capacitor discharge. A is correct.
US–China difference: Same parallel-branch equality in US theory.
Common pitfall: Treating the resistor voltage differently from the capacitor in a parallel circuit.
Real on-air practice: A bleed resistor across a cap sees the same decaying voltage.
将电阻R和电容C并联后接在电压为U的直流电源上。突然断开电源，流过电阻R的电流会：
- A从U/R按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到U/R
- C从0突然跳到U/R，然后再按指数规律逐渐减小到0
- D从U突然跳到0，然后再按指数规律逐渐增大到U/R
An R–C parallel circuit is connected to a DC supply of voltage U. When the supply is suddenly disconnected, the current through resistor R will:
- Afall from U/R to 0 exponentially
- Brise from 0 to U/R exponentially
- Cjump from 0 to U/R instantly, then fall to 0 exponentially
- Djump from U to 0 instantly, then rise to U/R exponentially
By Ohm’s law the resistor current equals its voltage over R; at disconnect it is U/R and decays as (U/R)·e^(−t/τ) to 0 with the shared voltage. A is correct.
US–China difference: Same relationship in US courses.
Common pitfall: Forgetting the initial current is U/R, not 0.
Real on-air practice: A bleeder resistor draws U/R initially when the supply is cut.
将电阻R和电容C并联后接到电压为U的直流电源上。突然断开电源，流过电容C的电流会：
- A从0突然跳到U/R，然后再按指数规律逐渐减小到0
- B从0突然跳到U/R并保持
- C从0按指数规律逐渐增加到U/R
- D从U突然跳到0，然后再按指数规律逐渐增大到U/R
An R–C parallel circuit is connected to a DC supply of voltage U. When the supply is suddenly disconnected, the current through capacitor C will:
- Ajump from 0 to U/R instantly, then fall to 0 exponentially
- Bjump from 0 to U/R instantly and stay there
- Crise from 0 to U/R exponentially
- Djump from U to 0 instantly, then rise to U/R exponentially
On disconnect the capacitor begins to discharge through R, so its current jumps from 0 (before disconnect there was no DC capacitor current) to U/R in the discharge direction and then decays exponentially to 0. A is correct — the magnitude follows the same (U/R)·e^(−t/τ) decay as the resistor current.
US–China difference: Same discharge-current behavior in US theory.
Common pitfall: Forgetting the capacitor current is zero before discharge then jumps at the switching instant.
Real on-air practice: A disconnect spark is the capacitor’s sudden discharge current.
将电阻R和电感L串联后突然接到直流电压U上，电感L两端的电压会：
- A从0突然跳到U，然后再按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到U
- C从U按指数规律逐渐减小到0
- D从U突然跳到0并保持
An R–L series circuit is suddenly connected to a DC voltage U; the voltage across inductor L will:
- Ajump from 0 to U instantly, then fall to 0 exponentially
- Brise from 0 to U exponentially
- Cfall from U to 0 exponentially
- Djump from U to 0 instantly and stay there
At t=0 the inductor current is zero (cannot change instantly), so the full U appears across L; as current builds, VL = U·e^(−t/τ) decays exponentially to 0. A is correct.
US–China difference: Same RL transient (VL starts at U) in US courses.
Common pitfall: Giving the capacitor’s rising curve to the inductor.
Real on-air practice: A relay coil shows a full supply kick at the moment of switch-on.
给定电阻R和电感L，则其阻值和电感量的乘积称为时间常数τ。将R和L串联后突然接到直流电压U上。经过τ时间后，电感L两端的电压大约为U的：
- A37%
- B99%
- C63%
- D6.28%
Given a resistor R and inductor L, the product of the resistance and inductance is the time constant τ. When R and L are connected in series and suddenly connected to DC voltage U, after time τ the voltage across inductor L is approximately what fraction of U?
- A37%
- B99%
- C63%
- D6.28%
After one τ the inductor voltage has decayed to e^(−1) ≈ 37% of its initial U (it started at U and falls toward 0). A is correct — the mirror of the RC capacitor’s 63% rise.
US–China difference: Same 37% decay value for RL in US material.
Common pitfall: Mixing up 37% (inductor decay) with 63% (resistor rise).
Real on-air practice: A flyback voltage across a coil drops to ~37% after one τ.
给定电阻R和电感L，则其阻值和电感量的乘积称为时间常数τ。将R和L串联后突然接到直流电压U上。经过τ时间后，电阻R两端的电压大约为U的：
- A63%
- B99%
- C37%
- D6.28%
Given a resistor R and inductor L, the product RL is the time constant τ. When connected in series and suddenly connected to DC voltage U, after time τ the voltage across resistor R is approximately what fraction of U?
- A63%
- B99%
- C37%
- D6.28%
The resistor voltage rises with the current: VR = U(1−e^(−t/τ)). After one τ it reaches 1−e^(−1) ≈ 63% of U. A is correct (the complement of the inductor’s 37% at that instant).
US–China difference: Same 63% rise for RL resistor in US courses.
Common pitfall: Swapping 63% (resistor) with 37% (inductor).
Real on-air practice: A series resistor’s drop climbs to ~63% of supply after one τ.
将电阻R和电感L串联后突然接到直流电压U上，电阻R两端的电压会：
- A从0按指数规律逐渐增加到U
- B从U按指数规律逐渐减小到0
- C从0突然跳到U，然后再按指数规律逐渐减小到0
- D从U突然跳到0并保持
When an R–L series circuit is suddenly connected to a DC voltage U, the voltage across resistor R will:
- Arise from 0 to U exponentially
- Bfall from U to 0 exponentially
- Cjump from 0 to U instantly, then fall to 0 exponentially
- Djump from U to 0 instantly and stay there
By Ohm’s law VR = I·R; the current starts at 0 and rises exponentially toward U/R, so VR rises from 0 toward U exponentially. A is correct; the inductor takes the initial U and gives it up.
US–China difference: Same RL charging curve in US theory.
Common pitfall: Assigning the inductor’s decaying curve to the resistor.
Real on-air practice: A dropping resistor’s voltage climbs as the inductor current builds.
将电阻R和电感L串联后突然接到直流电压U上，流过电阻R的电流会：
- A从0按指数规律逐渐增加到U/R
- B从0突然跳到U/R，然后再按指数规律逐渐减小到0
- C从U/R按指数规律逐渐减小到0
- D从0突然跳到U/R并保持
When an R–L series circuit is suddenly connected to a DC voltage U, the current through resistor R will:
- Arise from 0 to U/R exponentially
- Bjump from 0 to U/R instantly, then fall to 0 exponentially
- Cfall from U/R to 0 exponentially
- Djump from 0 to U/R instantly and stay there
Inductor current cannot jump, so the series current starts at 0 and rises as I = (U/R)(1−e^(−t/τ)) toward U/R. A is correct; the resistor current equals the inductor current.
US–China difference: Same RL current rise in US courses.
Common pitfall: Thinking the current jumps like a capacitor’s voltage — inductor current is continuous.
Real on-air practice: A keying line into an inductive load ramps up current gradually.
将电阻R和电感L串联后突然接到直流电压U上，流过电感L的电流会：
- A从0按指数规律逐渐增加到U/R
- B从0突然跳到U/R，然后再按指数规律逐渐减小到0
- C从U/R按指数规律逐渐减小到0
- D从0突然跳到U/R并保持
When an R–L series circuit is suddenly connected to a DC voltage U, the current through inductor L will:
- Arise from 0 to U/R exponentially
- Bjump from 0 to U/R instantly, then fall to 0 exponentially
- Cfall from U/R to 0 exponentially
- Djump from 0 to U/R instantly and stay there
Being a series circuit, the inductor current equals the resistor current: it starts at 0 and rises exponentially to U/R. A is correct; inductor current is continuous and cannot jump.
US–China difference: Same continuity-of-inductor-current rule in US theory.
Common pitfall: Letting the inductor current jump — only its voltage can jump.
Real on-air practice: An RF choke current eases up gradually when DC is applied.
将电阻R和电感L并联后接在电流为I的直流电路中。突然断开电路，电感L两端的电压会：
- A从I*R按指数规律逐渐减小到0
- B从0按指数规律逐渐增加到I*R
- C保持为I*R
- D始终为0
An R–L parallel circuit is placed in a DC circuit of current I. When the circuit is suddenly opened, the voltage across inductor L will:
- Afall from I·R to 0 exponentially
- Brise from 0 to I·R exponentially
- Cstay at I·R
- Dremain 0 at all times
On opening the source, the inductor maintains its current I, which now flows through the parallel resistor, so initially VL = I·R; this voltage then decays exponentially to 0 as the stored energy dissipates. A is correct.
US–China difference: Same inductor free-wheeling/decay in US theory; the spike can be large.
Common pitfall: Forgetting the inductor forces a momentary I·R voltage when the circuit opens.
Real on-air practice: A flyback diode protects transistors from this opening spike.
由电容与电感组成的串联或并联电路存在一种状态，此时电容的容抗与电感的感抗在特定频率下相互抵消，电路中的电抗消失。该状态称为：
- A谐振
- B匹配
- C负阻
- D幻象
A series or parallel circuit of a capacitor and an inductor has a state in which, at a particular frequency, the capacitive reactance and the inductive reactance cancel each other and the circuit’s reactance disappears. This state is called:
- Aresonance
- Bmatching
- Cnegative resistance
- Dphantom
谐振 (resonance) is the condition XL = Xc so the net reactance is zero. Matching, negative resistance, and phantom are unrelated concepts. A is correct.
US–China difference: Same definition of resonance in US material.
Common pitfall: Confusing resonance (reactance cancels) with impedance matching.
Real on-air practice: You tune your antenna system to resonance before matching to 50 Ω.
由电感为L的线圈和容量为C的电容组成的LC谐振电路的谐振频率f为：（“x＾m”表示“x的m次方”）
- Af = 1/（2π(LC)＾(1/2)）
- Bf = 2π(LC)＾(1/2)
- Cf = 1/((2πLC)＾2)
- Df = 2π(LC)＾(1/2)
The resonant frequency f of an LC resonant circuit made of a coil of inductance L and a capacitor of capacitance C is: (“x^m” means “x to the power m”)
- Af = 1 / (2π · (LC)^(1/2))
- Bf = 2π · (LC)^(1/2)
- Cf = 1 / ((2πLC)^2)
- Df = 2π · (LC)^(1/2)
The resonant frequency is f = 1/(2π√(LC)). A is correct; B/D give the reciprocal (proportional to √(LC)) and C has a squared denominator — all wrong.
US–China difference: Same f = 1/(2π√LC) in US exams.
Common pitfall: Dropping the 2π or inverting the formula.
Real on-air practice: You pick L and C to resonate your trap or filter at the band edge.
由电感为L的线圈和容量为C的电容组成的串联电路的总阻抗为：
- A谐振时呈现最小值
- B高于谐振频率呈现电感特性
- C低于谐振频率呈现电容特性
- D频率为零时呈现断路特性
The total impedance of a series circuit consisting of a coil of inductance L and a capacitor of capacitance C is: (Choose all that apply.)
- Aat resonance it is at a minimum
- Babove the resonant frequency it exhibits inductive behavior
- Cbelow the resonant frequency it exhibits capacitive behavior
- Dat zero frequency it behaves as an open circuit
For a series LC, Z = R + j(XL − Xc). At resonance XL = Xc so Z is minimum (A). Above resonance XL > Xc → inductive (B); below resonance Xc > XL → capacitive (C). At f = 0 the capacitor is an open circuit, so the series branch is open (D). All four are correct.
US–China difference: Same series-resonant impedance behavior in US theory.
Common pitfall: Thinking a series resonant circuit has maximum impedance (that is parallel).
Real on-air practice: A series-resonant trap passes its resonant frequency with least loss.
由电感为L的线圈和容量为C的电容组成的并联电路的总阻抗为：
- A谐振时呈现最大值
- B高于谐振频率呈现电容特性
- C低于谐振频率呈现电感特性
- D频率为零时呈现短路特性
The total impedance of a parallel circuit consisting of a coil of inductance L and a capacitor of capacitance C is: (Choose all that apply.)
- Aat resonance it is at a maximum
- Babove the resonant frequency it exhibits capacitive behavior
- Cbelow the resonant frequency it exhibits inductive behavior
- Dat zero frequency it behaves as a short circuit
For a parallel LC, impedance is maximum at resonance (A). Above resonance the inductor’s low reactance shunts the branch → net capacitive (B); below resonance the capacitor is the low-reactance path → net inductive (C). At f = 0 the coil is a short, so the parallel combination is a short (D). All four are correct.
US–China difference: Same parallel-resonant behavior in US theory (note the behaviors swap vs series).
Common pitfall: Applying series-resonance rules to a parallel tank.
Real on-air practice: A parallel tank is the high-impedance resonator in your PA output network.
我们通常讨论的谐振电路的Q值均指该电路与规定负载相连之后的“有载Q值”。但是，我们也时而讨论术语“空载Q值”。其涵义为：
- A空载Q值是回路自身储能与自身损耗之比，所以该值高代表回路的制作质量好
- B空载Q值主要取决于回路中电感的Q值，因为电容的损耗通常较电感为低
- C空载Q值越高则谐振电路的工作带宽越窄，所以没必要追求该值的最大化
- D空载Q值与有载Q值的比值过大说明回路的储能过多。这样的电路易打火损坏
The Q value of a resonant circuit we usually discuss refers to the “loaded Q” after the circuit is connected to a specified load. But we also sometimes discuss the term “unloaded Q”. Its meaning is: (Choose all that apply.)
- Athe unloaded Q is the ratio of the circuit’s own stored energy to its own loss, so a high value indicates good construction quality of the circuit
- Bthe unloaded Q mainly depends on the Q of the inductor in the circuit, because the capacitor’s loss is usually lower than that of the inductor
- Cthe higher the unloaded Q, the narrower the operating bandwidth of the resonant circuit, so there is no need to maximize this value
- Dtoo large a ratio of unloaded Q to loaded Q means the circuit stores too much energy; such a circuit is prone to arcing damage
空载Q值 (unloaded Q, Q₀) is the ratio of stored energy to inherent loss with no external load, reflecting coil/cap quality — A is correct. Since capacitor loss is small, Q₀ is dominated by the inductor’s Q — B is correct. C is false (a high Q₀ is generally desirable; it is the loaded Q that sets bandwidth), and D is false (the unloaded/loaded ratio is not an arcing criterion).
US–China difference: Same loaded vs unloaded Q distinction in US material.
Common pitfall: Blaming bandwidth on unloaded Q instead of loaded Q, or thinking Q₀ causes arcing.
Real on-air practice: A high-Q toroid in your tuner stores more energy and needs careful voltage rating.
根据谐振回路或者谐振天线电路Q值的高低，我们可以得到结论：
- AQ值越高，谐振曲线越尖锐，选择性越好，通带宽度越窄
- BQ值越高，谐振曲线越平坦，选择性越不明显，通带宽度越宽
- CQ值越高，回路或天线的工作状态受频率变化的影响越小
- DQ值越高，回路或天线的工作频率越高
From the magnitude of the Q of a resonant circuit or a resonant antenna circuit we can conclude:
- Athe higher the Q, the sharper the resonance curve, the better the selectivity, and the narrower the passband
- Bthe higher the Q, the flatter the resonance curve, the less obvious the selectivity, and the wider the passband
- Cthe higher the Q, the less the operating state of the circuit or antenna is affected by frequency changes
- Dthe higher the Q, the higher the operating frequency of the circuit or antenna
Quality factor Q = f₀/BW. A higher Q gives a sharper, more selective resonance curve and a narrower bandwidth (A). B describes low Q; C and D wrongly couple Q to frequency stability or operating frequency.
US–China difference: Same Q/BW relationship in US courses.
Common pitfall: Thinking high Q always “better” — it also narrows bandwidth, which can be undesirable.
Real on-air practice: A high-Q front-end filter rejects adjacent channels but passes a narrower slice.
谐振回路的通带宽度 (BW) 是指：
- A在回路谐振频率的两侧，信号衰减达到3dB时两个频率之间的间隔
- B在回路谐振频率的两侧，信号衰减达到30%时两个频率之间的间隔
- C在回路谐振频率的两侧，信号衰减达到80%时两个频率之间的间隔
- D在回路谐振频率的两侧，信号衰减达到95%时两个频率之间的间隔
The bandwidth (BW) of a resonant circuit is:
- Athe interval between the two frequencies on either side of the resonant frequency at which the signal attenuation reaches 3 dB
- Bthe interval between the two frequencies on either side of resonance at which attenuation reaches 30%
- Cthe interval between the two frequencies on either side of resonance at which attenuation reaches 80%
- Dthe interval between the two frequencies on either side of resonance at which attenuation reaches 95%
Bandwidth is defined at the −3 dB (half-power) points: the separation between the two frequencies where response falls by 3 dB from the peak. A is correct; the percentages in B/C/D are not the standard definition.
US–China difference: Same −3 dB (half-power) bandwidth definition in the US.
Common pitfall: Using a percentage drop instead of the standard 3 dB point.
Real on-air practice: Your receiver’s 2.7 kHz CW filter bandwidth is set between −3 dB points.
💬 Have questions about this topic, or FCC / CRAC exam preparation?
对本篇内容或 FCC / CRAC 备考有疑问？
本手册仅供业余无线电爱好者学习交流，题库原题版权归 CRAC（中国无线电协会业余无线电分会）所有，英文翻译由 BG7BAG 编译，转载请注明出处。
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