📚 Case Study: RLC Circuit Analysis Using Complex Numbers and Matrices | 案例分析实战演练：用复数和矩阵分析RLC电路
In this case study, we will apply several key concepts from CIE Year 12 Further Mathematics — complex numbers, matrices, differential equations, and phasor techniques — to analyse a practical AC circuit. Step by step, you will see how abstract mathematics provides powerful tools for solving real engineering problems.
在本案例分析中，我们将运用CIE Year 12进阶数学的核心概念——复数、矩阵、微分方程以及相量技术——来分析一个实际的交流电路。通过逐步操作，你将看到抽象的数学如何为解决实际工程问题提供强大的工具。
1. Introduction to the Case Study | 案例介绍
We consider a series-parallel RLC network driven by a sinusoidal voltage source. This type of circuit is commonly found in filters, tuners, and energy systems. The aim is to determine the steady-state currents and voltages using complex numbers and matrix algebra, mirroring the style of exam questions where multiple Further Pure topics are integrated.
我们考虑一个由正弦电压源驱动的串并联RLC网络。这类电路常见于滤波器、调谐器和能量系统。目标是利用复数和矩阵代数确定稳态电流和电压，这正反映了考试中综合多个进阶纯数主题的题型风格。
2. Physical Setup and Parameters | 物理设置与参数
The circuit consists of a voltage source vs(t) = Vm sin(ωt), a resistor R1 in series with a parallel combination of an inductor L and a series branch containing a resistor R2 and a capacitor C. All components are linear and time-invariant. We are given: Vm = 10 V, ω = 100 rad s−1, R1 = 5 Ω, R2 = 10 Ω, L = 0.1 H, C = 500 μF.
电路包含一个电压源 vs(t) = Vm sin(ωt)，一个电阻 R1 与一个由电感 L 和含电阻 R2 与电容 C 的串联支路组成的并联部分相串联。所有元件都是线性时不变的。已知：Vm = 10 V，ω = 100 rad s−1，R1 = 5 Ω，R2 = 10 Ω，L = 0.1 H，C = 500 μF。
3. Governing Differential Equation | 控制微分方程
Applying Kirchhoff’s voltage law to each loop yields integro-differential equations. For the loop through the inductor, the voltage across L is L(diL/dt). The capacitor voltage relates to current by vC = (1/C)∫ iC dt. These time-domain equations are coupled and difficult to solve directly. We therefore move to the frequency domain using complex exponentials.
对每个回路应用基尔霍夫电压定律会得到积分-微分方程。对经过电感的回路，L两端的电压为 L(diL/dt)。电容电压与电流的关系为 vC = (1/C)∫ iC dt。这些时域方程互相耦合，直接求解很困难。因此，我们将利用复指数转换到频域。
4. Phasor Transformation | 相量变换
Since the source is sinusoidal at a single frequency, all steady-state voltages and currents are sinusoids of the same frequency. We represent each sinusoidal quantity x(t) = Xm cos(ωt + φ) by a complex phasor X = Xm eiφ (using bold for phasors). The time derivative d/dt becomes multiplication by iω, and integration by 1/(iω). This transforms differential equations into algebraic equations.
由于电源是单一频率的正弦波，所有稳态电压和电流都是同频率的正弦量。我们将每个正弦量 x(t) = Xm cos(ωt + φ) 表示为一个复数相量 X = Xm eiφ（相量用粗体）。时间导数 d/dt 变为乘以 iω，积分变为乘以 1/(iω)。这就把微分方程转化成了代数方程。
5. Complex Impedance | 复数阻抗
Each passive element has a complex impedance: ZR = R, ZL = iωL, ZC = 1/(iωC) = −i/(ωC). Using these, we can treat the circuit as a resistive network with complex numbers, enormously simplifying the analysis. We calculate:
ωL = 100 × 0.1 = 10 Ω ⇒ ZL = i10 Ω
1/(ωC) = 1/(100 × 500×10−6) = 20 Ω ⇒ ZC = −i20 Ω
每个无源元件都有复数阻抗：ZR = R，ZL = iωL，ZC = 1/(iωC) = −i/(ωC)。利用它们，我们可以将电路视为含有复数的电阻网络，极大简化分析。我们计算：
ωL = 100 × 0.1 = 10 Ω ⇒ ZL = i10 Ω
1/(ωC) = 1/(100 × 500×10−6) = 20 Ω ⇒ ZC = −i20 Ω
6. Kirchhoff’s Laws in Phasor Form | 相量形式的基尔霍夫定律
We define mesh currents: I1 in the left mesh (through R1 and the parallel combination) and I2 in the right mesh (through L). The source phasor is Vs = 10∠0° = 10 + i0 V. Applying KVL and KCL in phasor form gives complex algebraic equations. The beauty of this method is that all AC circuit laws hold with impedances and phasors.
我们定义网孔电流：左网孔（通过R1和并联组合）的 I1，以及右网孔（通过L）的 I2。电源相量为 Vs = 10∠0° = 10 + i0 V。以相量形式应用KVL和KCL会得到复数代数方程。此方法的美妙之处在于，所有交流电路定律对阻抗和相量依然成立。
7. Matrix Formulation of Circuit Equations | 电路方程的矩阵表示
After writing the mesh equations and rearranging, we obtain a system of complex linear equations. The first mesh: Vs = (R1 + R2 + ZC)I1 − (R2 + ZC)I2. The second mesh: 0 = −(R2 + ZC)I1 + (R2 + ZC + ZL)I2. Substituting the impedances, we write this in matrix form Z I = V:
写出网孔方程并整理后，我们得到一个复数线性方程组。第一网孔：Vs = (R1 + R2 + ZC)I1 − (R2 + ZC)I2。第二网孔：0 = −(R2 + ZC)I1 + (R2 + ZC + ZL)I2。代入阻抗值，我们可将其写成矩阵形式 Z I = V：
[ 15−i20 −(10−i20) ] [ I1 ] = [ 10 ]
[ −(10−i20) 10−i10 ] [ I2 ] [ 0 ]
This is exactly the kind of matrix problem encountered in CIE Further Pure Mathematics, now with complex entries — a perfect demonstration of combined skills.
这正是CIE进阶纯数中会遇到的矩阵问题，只是现在矩阵元素是复数——这是对综合技能的完美演示。
8. Solving the Matrix Equation | 求解矩阵方程
We find the current phasors by inverting the 2×2 impedance matrix or using Cramer’s rule. The determinant Δ = (15−i20)(10−i10) − (−(10−i20))2. Compute each product:
(15−i20)(10−i10) = 150 − i150 − i200 + i2200 = 150 − i350 − 200 = −50 − i350.
(10−i20)2 = 100 − i400 + i2400 = 100 − i400 − 400 = −300 − i400.
Then Δ = (−50 − i350) − (−300 − i400) = 250 + i50. So Δ = 250 + i50.
通过对2×2阻抗矩阵求逆或使用克莱姆法则，我们可以求出电流相量。行列式 Δ = (15−i20)(10−i10) − (−(10−i20))2。计算每一项：
(15−i20)(10−i10) = 150 − i150 − i200 + i2200 = 150 − i350 − 200 = −50 − i350。
(10−i20)2 = 100 − i400 + i2400 = 100 − i400 − 400 = −300 − i400。
那么 Δ = (−50 − i350) − (−300 − i400) = 250 + i50。因此 Δ = 250 + i50。
Now, I1 = det([10, −(10−i20); 0, 10−i10]) / Δ = [10(10−i10) − (−(10−i20))·0] / Δ = (100 − i100) / (250 + i50). Simplify by multiplying numerator and denominator by the complex conjugate 250 − i50:
Numerator: (100 − i100)(250 − i50) = 25000 − i5000 − i25000 + i25000 = 25000 − i30000 − 5000 = 20000 − i30000.
Denominator: (250+i50)(250−i50) = 2502 + 502 = 62500 + 2500 = 65000.
Thus I1 = (20000 − i30000) / 65000 = 0.3077 − i0.4615 A.
现在，I1 = det([10, −(10−i20); 0, 10−i10]) / Δ = [10(10−i10) − 0] / Δ = (100 − i100) / (250 + i50)。通过分子分母同乘共轭复数 250 − i50 进行化简：
分子：(100 − i100)(250 − i50) = 25000 − i5000 − i25000 + i25000 = 25000 − i30000 − 5000 = 20000 − i30000。
分母：(250+i50)(250−i50) = 2502 + 502 = 62500 + 2500 = 65000。
因此 I1 = (20000 − i30000) / 65000 = 0.3077 − i0.4615 A。
Similarly, I2 = det([15−i20, 10; −(10−i20), 0]) / Δ = [0 − 10(−(10−i20))] / Δ = (100 − i200) / (250 + i50). After rationalising: numerator (100−i200)(250−i50) = 25000 − i5000 − i50000 + i210000 = 25000 − i55000 − 10000 = 15000 − i55000; denominator 65000 ⇒ I2 = 0.2308 − i0.8462 A.
同理，I2 = det([15−i20, 10; −(10−i20), 0]) / Δ = [0 − 10(−(10−i20))] / Δ = (100 − i200) / (250 + i50)。经有理化后：分子 (100−i200)(250−i50) = 25000 − i5000 − i50000 + i210000 = 25000 − i55000 − 10000 = 15000 − i55000；分母 65000 ⇒ I2 = 0.2308 − i0.8462 A。
9. Calculating Currents and Voltages | 计算电流和电压
We now have the mesh current phasors. The actual steady-state time-domain currents are found by taking the imaginary part (since the source was sin, not cos). The magnitude and phase of I1: |I1| = √(0.30772 + 0.46152) ≈ 0.555 A, angle φ1 = arctan(−0.4615/0.3077) ≈ −56.3°. So i1(t) ≈ 0.555 sin(100t − 56.3°) A. For I2: magnitude ≈ √(0.23082+0.84622) ≈ 0.877 A, φ2 ≈ arctan(−0.8462/0.2308) ≈ −74.7°, giving i2(t) ≈ 0.877 sin(100t − 74.7°) A. The branch currents and voltages across each element can be obtained by combining these phasors.
现在我们得到了网孔电流相量。实际稳态的时域电流可通过取虚部得到（因为电源是 sin 而非 cos）。I1 的幅值和相角：|I1| = √(0.30772 + 0.46152) ≈ 0.555 A，相角 φ1 = arctan(−0.4615/0.3077) ≈ −56.3°。因此 i1(t) ≈ 0.555 sin(100t − 56.3°) A。对于 I2：幅值 ≈ √(0.23082+0.84622) ≈ 0.877 A，φ2 ≈ arctan(−0.8462/0.2308) ≈ −74.7°，给出 i2(t) ≈ 0.877 sin(100t − 74.7°) A。各支路电流和各元件电压可以通过组合这些相量得出。
10. Interpreting Phase Angles and Resonance | 解释相位角与谐振
The negative phase angles indicate that both currents lag the source voltage, as expected in an inductive-capacitive network. Notice that I2 (through the inductor) has a larger lag than I1, reflecting the inductive nature of its path. By adjusting ω, we could achieve resonance where the imaginary part of the total impedance vanishes. This case study reinforces how complex numbers encode both magnitude and phase in a single algebraic entity, a central idea in Further Mathematics.
负的相角表明两个电流均滞后于电源电压，这在电感-电容网络中符合预期。注意到通过电感的 I2 比 I1 的滞后更大，反映了该通路的感性本质。通过调整 ω，我们可以使总阻抗的虚部为零，从而实现谐振。本案例强化了复数如何在单一代数实体中同时编码幅值和相位这一进阶数学的核心思想。
11. Practical Considerations and Exam Tips | 实际考虑与考试技巧
- Always convert all quantities to phasors and impedances before writing equations — this avoids calculus completely.
- When solving complex matrix equations, keep order of operations clear: determinant first, then divide each cofactor by the determinant.
- Check your phasor-to-time conversion: if source is sin, take Im part; if cos, take Re part.
- Use a consistent unit system (Ω, H, F, s) and verify that all imaginary parts have the correct sign.
- In the exam, marks are often awarded for correct method even if a numerical slip occurs; show all workings neatly.
- 在写方程前，始终将所有量转换为相量和阻抗——这完全避开了微积分。
- 解复矩阵方程时，保持运算顺序清晰：先算行列式，再用每个余子式除以行列式。
- 检查相量到时域的转换：若电源为 sin，取虚部；若为 cos，取实部。
- 使用一致的单位制（Ω、H、F、s），并验证所有虚部的符号正确。
- 在考试中，即使出现数值错误，正确的方法通常也能得分；清晰地展示所有步骤。
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