# 计算交流电路中的等效阻抗

   

## 带有解决方案的例子

$$Z_1$$和$$Z_2$$串联，等效阻抗$$Z_{AB}$$由串联阻抗规则给出：
$$Z_{AB} = Z_1 + Z_2$$

$$\dfrac{1}{Z_2} = \dfrac{1}{j\omega L} + \dfrac{1}{\dfrac{1}{ j \omega C}}$$

$$\dfrac{1}{Z_2} = \dfrac{1}{j\omega L} + j \omega C$$

$$\dfrac{1}{Z_2} = \dfrac{1-\omega^2 C L}{j\omega L}$$

$$Z_2 = \dfrac{j\omega L}{ 1-\omega^2 C L}$$

$$Z_{AB} = R + \dfrac{j\omega L}{ 1-\omega^2 C L}$$

$$|Z_{AB}| = \sqrt {R^2 + \left(\dfrac{\omega L}{ 1-\omega^2 C L )} \right)^2 }$$
$$\theta = \arctan \dfrac{\omega L}{R(1-\omega^2 C L)}$$

$$Z_{AB} = \sqrt {R^2 + \left(\dfrac{\omega L}{ 1-\omega^2 C L )} \right)^2 } e^{\arctan \dfrac{\omega L}{R(1-\omega^2 C L)}}$$

$$Z_{AB} = \sqrt {R^2 + \left(\dfrac{\omega L}{ 1-\omega^2 C L )} \right)^2 } \; \angle \; {\arctan \dfrac{\omega L}{R(1-\omega^2 C L)}}$$

$$L_1 = 20 \; mH$$ , $$C_1 = 10 \; \mu F$$ , $$L_2 = 40 \; mH$$ , $$C_2 = 30 \; \mu F$$ 信号频率 $$f = 1.5 \; kHz$$

$$Z_1$$和$$Z_2$$串联，如下图所示，因此使用串联阻抗规则计算$$Z_{AB}$$如下：
$$Z_{AB} = Z_1 + Z_2$$

$$\dfrac{1}{Z_1} = \dfrac{1}{j\omega L_1} + \dfrac{1}{\dfrac{1}{ j \omega C_1}}$$

$$\dfrac{1}{Z_1} = \dfrac{1}{j\omega L_1} + j \omega C_1$$

$$\dfrac{1}{Z_1} = \dfrac{1 - \omega^2 L_1 C_1}{j\omega L_1}$$

$$Z_1 = \dfrac{j\omega L_1}{1 - \omega^2 L_1 C_1}$$
$$Z_2$$可以类似于$$Z_1$$的方式计算，结果为：
$$Z_2 = \dfrac{j\omega L_2}{1 - \omega^2 L_2 C_2}$$

$$Z_{AB} = \dfrac{j\omega L_1}{1 - \omega^2 L_1 C_1} + \dfrac{j\omega L_2}{1 - \omega^2 L_2 C_2}$$

$$Z_{AB} = j \omega \left (\dfrac{ L_1}{1 - \omega^2 L_1 C_1} + \dfrac{ L_2}{1 - \omega^2 L_2 C_2} \right)$$

$$Z_{AB} \approx - j 14.81$$

$$|Z_{AB}| \approx 14.81$$
$$\theta = - \pi / 2$$

$$Z \approx 14.81 \; e^{-j \pi/2}$$

$$Z \approx 14.81 \angle - \pi/2$$

$$R_1 = 20 \; \Omega$$ , $$C_1 = 50 \; \mu F$$ , $$C_2 = 40 \; \mu F$$ , $$R_2 = 80 \; \Omega$$ 信号频率 $$f = 0.5 \; kHz$$

$$Z_1 = R_1$$
$$Z_2 = \dfrac{1}{j \omega C_1}$$
$$Z_3 = R_2 + \dfrac{1}{j \omega C_2}$$
$$Z_2$$ 和 $$Z_3$$ 并联，它们的等效阻抗 $$Z_{2,3}$$ 使用并联阻抗规则如下：
$$\dfrac{1}{Z_{2,3}} = \dfrac{1}{Z_2} + \dfrac{1}{Z_3}$$
$$Z_{2,3} = \dfrac{Z_2 \cdot Z_3}{Z_2 + Z_3}$$

$$Z_1$$ 和 $$Z_{2,3}$$ 串联，因此
$$Z_{AB} = Z_1 + Z_{2,3} = Z_1 + \dfrac{Z_2 \cdot Z_3}{Z_2 + Z_3}$$

$$Z_{AB} = R_1 + \dfrac{\dfrac{1}{j \omega C_1} \cdot (R_2 + \dfrac{1}{j \omega C_2})}{\dfrac{1}{j \omega C_1} + R_2 + \dfrac{1}{j \omega C_2}}$$

$$Z_{AB} \approx 20.49 -6.29 j$$
$$Z_{AB}$$ 的模：
$$| Z_{AB} | \approx \sqrt{20.49^2 + (-6.29)^2 } = 21.43$$
$$Z_{AB}$$ 的辐角：
$$\theta \approx \arctan (\dfrac{-6.29}{20.49}) = -0.20 rad$$ 或 $$\theta = -17.07^{\circ}$$

$$Z_{AB} \approx 21.43 e^{ -0.20 j}$$

$$Z_{AB} \approx 21.43 \angle -17.07^{\circ}$$