CRAC Class B (2025) — Questions 1061–1070

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U1128. The output signal voltage of a circuit is 10,000 times the input signal voltage. The gain of this circuit is: ("x^m" means "x to the power m")

某电路输出信号电压是输入信号电压的1万倍。该电路的增益 为：（“x＾m”表示“x的m次方”）

Exam tips · 考试提示

Voltage gain = 20·log₁₀(10⁴) = 20×4 = 80 dB. Each factor of 10 in voltage is +20 dB, so 10⁴ → 80 dB. B/C/D wrongly treat the numeric value as dB.

U1129. The output signal voltage of a circuit is 10 times the input signal voltage. The gain of this circuit is approximately:

某电路输出信号电压是输入信号电压的10倍。该电路的增益约 为：

Exam tips · 考试提示

Voltage gain = 20·log₁₀(10) = 20 dB. (7 dB is the power equivalent of 5×; 14 dB would be 20·log₁₀(5).)

U1130. The input signal voltage of a circuit is half the output signal voltage. The gain of this circuit is approximately:

某电路输入信号电压是输出信号电压的一半。该电路的增益约 为：

Exam tips · 考试提示

Input = ½ output ⇒ output = 2× input. Voltage gain = 20·log₁₀(2) ≈ 6.02 ≈ 6 dB (positive gain). B is the inverse; C/D are power-domain values.

U1131. The output signal power of a circuit is 1/100 of the input signal power. The gain of this circuit is:

某电路输出信号功率是输入信号功率的1/100。该电路的增益 为：

Exam tips · 考试提示

Gain = 10·log₁₀(1/100) = 10×(−2) = −20 dB (a 100× power attenuation). B/C/D misapply the sign or factor.

U1132. The output signal power of a circuit is one millionth of the input signal power. The gain of this circuit is:

某电路输出信号功率是输入信号功率的百万分之一。该电路的 增益为：

Exam tips · 考试提示

Gain = 10·log₁₀(10⁻⁶) = 10×(−6) = −60 dB. Each 10× power reduction is −10 dB, so 10⁶ → −60 dB. C/D are nonsensical literal conversions.

U1133. The output signal power of a circuit is 1/5 of the input signal power. The gain of this circuit is approximately:

某电路输出信号功率是输入信号功率的1/5。该电路的增益约 为：

Exam tips · 考试提示

Gain = 10·log₁₀(1/5) = −10·log₁₀(5) ≈ −6.99 ≈ −7 dB. (The +7 dB from U1125 is the inverse; here it's a loss, so negative.)

U1134. The output signal power of a circuit is half the input signal power. The gain of this circuit is approximately:

某电路输出信号功率是输入信号功率的一半。该电路的增益约 为：

Exam tips · 考试提示

Gain = 10·log₁₀(0.5) ≈ −3.01 ≈ −3 dB. This is the famous "half power = −3 dB" relation. B is the inverse (a doubling).

U1135. The output signal voltage of a circuit is 1/100 of the input signal voltage. The gain of this circuit is:

某电路输出信号电压是输入信号电压的1/100。该电路的增益 为：

Exam tips · 考试提示

Voltage gain = 20·log₁₀(1/100) = 20×(−2) = −40 dB. (Compare U1127: 100× → +40 dB; here it's the inverse.)

U1136. The output signal voltage of a circuit is one ten-thousandth of the input signal voltage. The gain of this circuit is: ("x^m" means "x to the power m")

某电路输出信号电压是输入信号电压的万分之一。该电路的增 益为：（“x＾m”表示“x的m次方”）

Exam tips · 考试提示

Voltage gain = 20·log₁₀(10⁻⁴) = 20×(−4) = −80 dB. Each 10× voltage reduction is −20 dB, so 10⁻⁴ → −80 dB. B/C/D are wrong literal conversions.

U1137. The output signal voltage of a circuit is 1/10 of the input signal voltage. The gain of this circuit is approximately:

某电路输出信号电压是输入信号电压的1/10。该电路的增益约 为：

Exam tips · 考试提示

Voltage gain = 20·log₁₀(1/10) = −20 dB. (The inverse of U1129.) B is the power-domain 5× loss value.