# 串联RLC电路中的电流和电压计算

   

## A - 串联RLC电路中的复数阻抗和相量

$$V_i - V_R - V_L - V_C = 0$$     (1)

$$V_R = Z_R I$$
$$V_L = Z_L I$$
$$V_C = Z_C I$$

$$V_i = Z_R I + Z_L I + Z_C I = 0$$

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

$Z = Z_R + Z_L + Z_C = R + j \left(\omega L - \dfrac{1}{\omega C} \right)$
$$Z$$ 的模数为 $|Z| = \sqrt{R^2 + \left(\omega L - \dfrac{1}{\omega C}\right)^2}$
$$Z$$ 的幅角为 $\theta = arctan \left( \dfrac {\omega L - \dfrac{1}{\omega C}}{R} \right)$

$Z = |Z| \; \angle \; \theta$

## B - 串联RLC电路的电流和电压

$$e^{j \omega t} = \cos (\omega t) + j \sin (\omega t )$$

$$v_i$$ 等于 $$e^{j \omega t}$$ 的实部。

$$V_i = V_0 e^{j \omega t}$$

$$I = \frac{V_0 e^{j\omega t}}{|Z| \; \angle \; \theta}$$

$$I = \dfrac{V_0}{|Z|} \; \angle \; \omega t - \theta$$
$$I = I_0 \; \angle \; \omega t - \theta$$，其中 $$I_0 = \dfrac{V_0}{|Z|}$$

$$Z_R = R = R \; \angle \; 0$$
$$Z_L = j \omega L = \omega L \; \angle \; \pi/2$$
$$Z_C = - \dfrac{1}{\omega C} j = \dfrac{1}{\omega C} \; \angle \; - \pi/2\$$

$$V_R = Z_R I = (R \; \angle \; 0) (I_0 \; \angle \; {\omega t - \theta}) = R I_0 \angle \; \omega t - \theta$$
$$V_L = Z_L I = (\omega L \; \angle \; \pi/2) (I_0 \; \angle \; {j\omega t - \theta}) = \omega L I_0 \; \angle \; {\omega t - \theta + \pi/2}$$
$$V_C = Z_C I = (\dfrac{1}{\omega C} \; \angle \; - \pi/2) (I_0 \; \angle \; {j\omega t - \theta}) = \dfrac{I_0}{\omega C} \; \angle \; {\omega t - \theta - \pi/2}$$

$$i = \dfrac{V_0}{|Z|} \cos( \omega t - \theta)$$

$$v_R = R \dfrac{V_0}{|Z|} \cos(\omega t - \theta)$$

$$v_L = \omega L \dfrac{V_0}{|Z|} \cos(\omega t - \theta + \pi/2)$$

$$v_C = \dfrac{V_0}{\omega C|Z|} \cos(\omega t - \theta - \pi/2)$$

## C - 带有详细解答的示例

a) 计算电容、电感和电阻的阻抗以及串联RLC电路的等效阻抗 $$Z$$ 的复数形式。
b) 计算电流和所有电压的复数形式。
c) 计算实际电流和电压。

a)

$$Z_R = R = 400 \; \Omega$$

$$Z_L = j \omega L = j \cdot 1000 \cdot 400 \cdot 10^{-3} = 400 j \; \Omega$$

$$Z_C = - \dfrac{1}{\omega C} j = - \dfrac{1}{1000 \cdot 200 \cdot 10^-6} j = - 5 j \; \Omega$$
$$Z = Z_R + Z_L + Z_C = 400 + 400 j - 5 j = 400 + 395 j$$
b)
$$v_i = 20 cos ( \omega t)$$，因此电源电压的极坐标形式 $$V_i = 20 \; \angle \; 0$$

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

$$I = \dfrac{20 \; \angle \; 0}{400 + 395 j}$$

$$400 + 395 j = \sqrt {400^2+395^2} \; \angle \; \arctan\left(\dfrac{395}{400}\right) = 562.16 \; \angle \; 44.64^{\circ}$$

$$I = \dfrac{20 \; \angle \; 0}{562.16 \; \angle \; 44.64^{\circ}} = \dfrac{20}{562.16} \; \angle \; 0 - 44.64^{\circ}$$

$$I = 0.0356 \; \; \angle \; - 44.64^{\circ}$$ A
$$V_R = R I = 400 (0.0356 \; \; \angle \; - 44.64^{\circ}) = 14.24 \; V \; \angle \; - 44.64^{\circ}$$ V
$$V_L = Z_L I = 400 j (0.0356 \; \; \angle \; - 44.64^{\circ}) = 14.24 \; V \; \angle \; 45.36^{\circ}$$ V
$$V_C = Z_C I = - 5 j (0.0356 \; \; \angle \; - 44.64^{\circ}) = 0.18 \; V \; \angle \; -134.6^{\circ}$$ V

a) 计算阻抗 $$Z$$ 的虚部为零时的角频率 $$\omega$$。
b) 计算在部分a中找到的频率下的电流和电压。

a)

$$\omega L - \dfrac{1}{\omega C} = 0$$

$$\omega^2 L C = 1$$
$$\omega = \dfrac{1}{\sqrt{L C}}$$

$$\omega = \dfrac{1}{ \sqrt{ 200\cdot10^{-3} \cdot 200 \cdot 10^{-6}}} = 158.11$$ rad/s
b)
$$I = \dfrac{V_i}{Z} = \dfrac{10 \; \angle \; 0}{R \; \angle \; 0} = \dfrac{10}{500} \; \angle \; 0 = 0.02 \; \angle\; 0$$
$$V_R = R I = 500 \cdot 0.02 \; \angle \; 0 = 10 \; \angle\; 0$$
$$V_C = Z_C I = \dfrac{1}{\omega C} \; \angle \; -90^{\circ} \cdot 0.02 \; \angle\; 0 = \dfrac{1}{\omega C} \cdot 0.02 \; \angle \; -90^{\circ} = \dfrac{1}{158.11 \cdot 200 \cdot 10^{-6}} \cdot 0.02 \; \angle \; -90^{\circ} = 0.6324 \; \angle \; -90^{\circ}$$
$$V_L = Z_L I = \omega L \; \angle \; 90^{\circ} \cdot 0.02 \; \angle\; 0 = \omega L \cdot 0.02 \; \angle \; 90^{\circ} = 158.11 \cdot 200 10^{-3} \cdot 0.02 \; \angle \; 90^{\circ} = 0.6324 \; \angle \; 90^{\circ}$$

a) 计算电容、电感和电阻串联时的总阻抗 $$Z$$，并写成极坐标形式 $$Z = |Z| \; \angle \; \theta$$
b) 求使 $$\theta = -60^{\circ}$$ 的频率 $$f$$。

a)

$$Z = R + j(\omega L - \dfrac{1}{\omega C}) = 200 + j( \omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}})$$
$$Z = \sqrt {200^2 + ( \omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}})^2} \; \angle \; \arctan \left(\dfrac{\omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}}}{200}\right)$$
b)
$$\arctan \left(\dfrac{\omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}}}{200}\right) = -60^{\circ}$$
$$\dfrac{\omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}}}{200} = \tan (-60^{\circ}) = -1.73205$$

$$\omega \cdot 100 \cdot 10^{-3} - \dfrac{1}{\omega \cdot 47 \cdot 10^{-6}} = -346.41$$

$$47 \cdot 10^{-7} \omega^2 - 1 = -0.01628127 \omega$$

$$\omega = 53.23$$ rad/s
$$\omega = 2 \pi f = 60.3682$$
$$f = \dfrac{60.3682}{2 \pi} = 9.60789$$ Hz

a) 计算电容、电感和电阻串联时的总阻抗 $$Z$$，并写成极坐标形式 $$Z = |Z| \; \angle \; \theta$$
b) 求使 $$\theta = 40^{\circ}$$ 并且 $$|Z| = 100$$ 的电阻 $$R$$ 和频率 $$f$$。

a)

$$Z = \sqrt{R^2 + (\omega L - \dfrac{1}{\omega C})^2} \; \angle \; \arctan \left( \dfrac{\omega L - \dfrac{1}{\omega C}}{R} \right)$$
b)
$$\theta = \arctan \left( \dfrac{\omega L - \dfrac{1}{\omega C}}{R} \right) = 40^{\circ}$$
$$\dfrac{\omega L - \dfrac{1}{\omega C}}{R} = \tan 40^{\circ} = 0.83909$$
$$\omega L - \dfrac{1}{\omega C} = 0.83909 R$$

$$\sqrt{R^2 + (0.83909 R)^2} = 100$$

$$R = 76.6048 \; \Omega$$
$$\omega L - \dfrac{1}{\omega C} = 0.83909 R = 64.27832$$

$$\omega = 1317.8557$$ rad/s
$$f = \dfrac{1317.8557}{2\pi} = 209.74324$$ Hz