# 用解决方案解决交流电路问题

   

## 复数的回顾

$$Z = r \; \angle \; \theta$$

$$r = |Z| = \sqrt {a^2 + b^2}$$ 和 $$\theta = \arctan \left( \dfrac{b}{a} \right)$$，范围在 $$-\pi \lt \theta \le \pi$$

$$Z_1 \cdot Z_2 = r_1 \cdot r_2 \; \angle \; \theta_1 + \theta_2$$
$$\dfrac{Z_1}{Z_2} = \dfrac{r_1}{r_2} \; \angle \; \theta_1 - \theta_2$$



## 带解决方案的问题

$$V_i - V_R - V_C = 0$$      (1)

$$Z_R = R \;$$       (实数)
$$Z_C = - j \dfrac{1}{\omega C} \;$$       (虚数)

$$V_R = Z_R I$$ 和 $$V_C = Z_C I$$

$$V_i - Z_R I - Z_C I = 0$$

$$I = \dfrac{V_i}{Z_R + Z_C}$$

$$V_R = Z_R I = \dfrac{V_i Z_R }{Z_R + Z_C}$$

$$V_C = Z_C I = \dfrac{V_i Z_C }{Z_R + Z_C}$$

$$V_i = 10 \; \angle \; 0$$

$$I = \dfrac{10 \; \angle \; 0}{R - j \dfrac{1}{\omega C}}$$

$$V_R = \dfrac{10 R \; \angle \; 0 \; }{R - j \dfrac{1}{\omega C}}$$

$$V_C = \dfrac{ (10 \; \angle \; 0 \;) (- j \dfrac{1}{\omega C}) }{R - j \dfrac{1}{\omega C}}$$

$$Z_D$$ 的模为：$$| Z_D | = \sqrt {R^2 + \dfrac{1}{\omega^2 C^2}}$$

$$Z_D$$ 的辐角：$$\phi = \arctan \dfrac{-\dfrac{1}{\omega C}}{R} = \arctan \dfrac{-1}{R \omega C}$$

$$- j \dfrac{1}{\omega C} = \dfrac{1}{\omega C} \; \angle \; -\dfrac{\pi}{2}$$

$$I = \dfrac{10 \; \angle \; 0}{ | Z_D | \; \angle \; \phi}$$

$$V_R = \dfrac{10 R \; \angle \; 0 \; }{ | Z_D | \; \angle \; \phi}$$

$$V_C = \dfrac{10 \; \angle \; 0 \; (\dfrac{1}{\omega C} \; \angle \; -\dfrac{\pi}{2}) }{ | Z_D | \; \angle \; \phi}$$

$$I = \dfrac{10} { | Z_D | } \; \angle \; - \phi$$

$$V_R = \dfrac{10 R}{ | Z_D |} \; \angle \; - \phi$$

$$V_C = \dfrac{10}{ \omega C | Z_D | } \; \angle \; -\dfrac{\pi}{2}- \phi$$

$$\omega = 2 \cdot \pi \cdot f = 2 \cdot \pi 10^3 = 2000 \pi$$
$$| Z_D | = \sqrt {R^2 + \dfrac{1}{\omega^2 C^2}} = \sqrt {100^2 + \dfrac{1}{(2000 \pi)^2 (0.47 \cdot 10^{-6})^2}} = 353.08 \; \Omega$$
$$\phi = \arctan \dfrac{-\dfrac{1}{\omega C}}{R} = \arctan \dfrac{-\dfrac{1}{2000 \pi \cdot 0.47 \cdot 10^{-6} }}{100} = -73.55^{\circ}$$

$$I = \dfrac{10} { | Z_D | } \; \angle \; - \phi = 0.0283 \; \angle \; 73.55^{\circ}$$
$$I$$ 的模为 $$0.02832 \; A$$ ，相移为 $$73.55^{\circ}$$

$$V_R = \dfrac{10 R}{ | Z_D |} \; \angle \; - \phi = 2.832 \; \angle \; 73.55^{\circ}$$
$$V_R$$ 的模为 $$2.832 \; V$$ ，相移为 $$73.55^{\circ}$$

$$V_C = \dfrac{10}{ \omega C | Z_D | } \; \angle \; -\dfrac{\pi}{2}- \phi = 9.591 \; \angle \; -16.45^{\circ}$$
$$V_C$$ 的模为 $$9.591 \; V$$ ，相移为 $$-16.45^{\circ}$$

$$V_i$$ 可以写成极坐标形式如下
$$V_i = 10 \; \angle \; 0$$

$$Z_R = R \;$$

$$Z_C = - j \dfrac{1}{\omega C} \;$$

$$Z = \dfrac{Z_R \cdot Z_C}{Z_R + Z_C}$$

$$Z_L = j \omega L$$

$$V_i - V_L - V_0 = 0$$      (1)

$$V_L = Z_L I$$ 和 $$V_0 = Z I$$

$$V_i - Z_L I - Z I = 0$$

$$I = \dfrac{V_i}{ Z_L + Z}$$

$$V_o = Z I = \dfrac{Z}{ Z_L + Z} V_i$$

$$\omega = 2 \pi f = 2 \pi \cdot 2000 = 4000 \pi$$
$$Z_R = R = 100$$
$$Z_C = - j \dfrac{1}{\omega C} = - j \dfrac{1}{4000 \pi \cdot 0.47 \cdot 10^{-6}} = - 169.314 j$$

$$Z = \dfrac{R \cdot Z_C}{R + Z_C}$$ ( ($$R$$ 和 $$C$$ 并联)

$$= \dfrac{100 \cdot (- 169.314 j)}{100 - 169.314 j } = 74.138 -43.787 j$$

$$Z_L = j \omega L = 4000 \pi \cdot 300 \cdot 10^{-3} j = 3769.911 j$$
$$V_o = \dfrac{Z}{ Z_L + Z} V_i$$

$$= \dfrac{Z}{ Z_L + Z} V_i$$

$$= \dfrac{74.138 -43.787 j}{ 3769.911 j + 74.138 -43.787 j} V_i$$

$$= (-0.01135 -0.02012 j) V_i$$

$$= ( 0.02310 \angle -119.43^{\circ} ) (10 \; \angle \; 0)$$

$$V_o = 0.23 \; \angle \; -119.43^{\circ}$$

$$V_i$$ 可以写成极坐标形式如下
$$V_i = 10 \; \angle \; 0$$
$$V_o$$ 可以通过欧姆定律计算
$$V_o = R_3 I_3$$

$$Z_1 = R_1$$
$$Z_2 = \dfrac{R_2 ( - \dfrac{1}{\omega C_1} j)}{R_2 - \dfrac{1}{\omega C_1} j}$$      ( $$C_1$$ 和 $$R_2$$ 并联)
$$z_3 = R_3 + j \omega L - \dfrac{1}{ \omega C_2} j$$      ( $$C_2$$、$$L$$ 和 $$R_3$$ 串联)

$$V_i - V_{z_1} - V_{z_2} = 0$$
$$V_{z_2} - V_{z_3} = 0$$

$$V_i - Z_1 I_1 - Z_2 I_2 = 0$$     (1)
$$Z_2 I_2 - Z_3 I_3 = 0$$     (2)

$$I_1 = I_2 + I_3$$     (3)

$$V_i - Z_1 ( I_2 + I_3) - Z_2 I_2 = 0$$     (4)
$$Z_2 I_2 - Z_3 I_3 = 0$$     (5)

$$(Z_1 + Z_2 ) I_2 + Z_1 I_3 = V_i$$     (4)
$$Z_2 I_2 - Z_3 I_3 = 0$$     (5)

$$I_3 = \dfrac{\begin{vmatrix} Z_1 + Z_2 & V_i \\ Z_2 & 0 \end{vmatrix}}{\begin{vmatrix} Z_1 + Z_2 & + Z_1 \\ Z_2 & -Z_3 \end{vmatrix}}$$

$$I_3 = \dfrac{Z_2}{(Z_1+Z_2)Z_3 + Z_1 Z_2} V_i$$

$$\omega = 2 \pi f = 5000 \pi$$
$$Z_1 = 220$$
$$Z_2 = \dfrac{2200 ( - \dfrac{1}{5000 \pi \cdot 0.47 10^{-6}} j)}{2200 - \dfrac{1}{5000 \pi \cdot 0.47 10^{-6}} j}$$

$$Z_2 = 8.30804 -134.93950 j$$

$$z_3 = 1000 + 5000 \pi \cdot 30 \cdot 10^{-3} j - \dfrac{1}{ 5000 \pi \cdot 1.5 \cdot 10^{-6}} j$$
$$Z_3 = 1000 + 428.79757 j$$
$$I_3 = (0.00013 - 0.00043 j)V_i = 0.00044 \angle -73.18^{\circ} \cdot 10 \angle 0$$
$$V_0 = R_3 I_3 = 1000 \cdot 0.00044 \angle -73.18^{\circ} \cdot 10 \angle 0$$

$$V_0 = 4.4 \angle -73.18^{\circ} V$$