# 交流电路中的功率

   

## A - 交流电路中的瞬时功率

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

$P(t) = i(t) \; v(t) = \dfrac{V_0^2}{|Z|} \; \cos ( \omega t - \theta) \; \cos(\omega t)$

## B - 交流电路中的平均功率

$P_a = \displaystyle \dfrac{1}{T} \int_0^T P(t) dt$

$$P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; \cos ( \omega t - \theta) \; \cos(\omega t) \; dt$$

$$P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; (\cos^2 \omega t \; \cos \theta + \sin \omega t \; \cos \omega t \; \sin \theta ) \; dt$$

$$P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; cos^2 \omega t \; \cos \theta \; dt + \dfrac{V_0^2}{T |Z|} \int_0^T \; \sin \omega t \; \cos \omega t \; \sin \theta \; dt$$

$$\displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; \sin \omega t \; \cos \omega t \; \sin \theta \; dt = \dfrac{V_0^2}{2 T |Z|} \sin \theta) \int_0^T \sin (2 \omega t ) \; dt$$
$$\quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } \left[\cos (2 \omega t ) \right]_0^T$$
$$\quad \quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } \left[\cos 2 \omega T - \cos 0 \right]$$

$$\quad \quad \quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } [\cos (4 \pi) - \cos 0]$$
$$\quad \quad \quad \quad = 0$$

$$P_a = \displaystyle \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; (\cos(2 \omega t )+1) \; dt$$

$$P_a = \displaystyle \dfrac{V_0^2}{2 T |Z|} \cos \theta \int_0^T \; \dfrac{1}{2} \cos(2 \omega t ) \; dt + \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; dt$$

$$\displaystyle P_a = \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; dt$$
$$\displaystyle \quad = \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \left[t\right]_0^T$$
$\displaystyle \quad \quad P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta$

a) 求出串联RLC电路的总阻抗 $$Z$$，并将其表示为极坐标形式。
b) 求出传递到总阻抗 $$Z$$ 的平均功率。

a)

$$\omega = 2 \pi f = 4000 \pi$$ 弧度/秒

$$Z = 1000 + j\left(4000 \pi \times 100 \times 10^{-3} - \dfrac{1}{4000 \pi \times 100 \times 10^{-6}} \right)$$
$$Z = 1000 + \left(400\pi -\dfrac{5}{2\pi} \right) j$$

$$\theta = \arctan \left(\dfrac{400\pi -\dfrac{5}{2\pi} }{1000} \right)$$
$$|Z| = \sqrt {1000^2 + \left(400\pi -\dfrac{5}{2\pi}\right)^2}$$

b)
$$\displaystyle P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta$$

$$\displaystyle P_a = \dfrac{5^2}{2 \sqrt {1000^2 + \left(400\pi -\dfrac{5}{2\pi}\right)^2} } \cos \left( \arctan \left(\dfrac{400\pi -\dfrac{5}{2\pi} }{1000} \right) \right)$$
$$\quad \approx 0.00485 \; 瓦$$

a) 对于串联RLC电路，总阻抗公式为：  $$Z = R + j(\omega L - \dfrac{1}{\omega C})$$

$$Z = R + j (\dfrac{1}{\sqrt{LC}} L - \dfrac{1}{\dfrac{C}{\sqrt{LC}}})$$
$$\quad = R + j ( \dfrac{1}{\sqrt{LC}} L - \dfrac{\sqrt{LC}}{C} )$$
$$\quad = R + j ( \dfrac{\sqrt L}{\sqrt C} - \dfrac{\sqrt{L}}{\sqrt C} )$$

$$Z = R$$

$$|Z| = R$$

$$P_a max = \dfrac{V_0^2}{2 R}$$

a) 通过上面的公式表达平均功率 $$P_a$$，绘制 $$P_a$$ 随角频率 $$\omega$$ 变化的图，并找出 $$P_a$$ 最大值的位置。
b) 验证功率在角频率 $$\omega_r = \dfrac{1}{\sqrt{LC}}$$（或频率 $$f_r = \dfrac{1}{2 \pi \sqrt{LC}}$$）处达到最大值，并且如示例2中解释的那样，由 $$P_a max = \dfrac{V_0^2}{2 R}$$ 给出。

$P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta$

$$P_a (\omega ) = \dfrac{V_0^2}{2 \sqrt { R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}} \cos \left(\arctan \left( \dfrac{\omega L - \dfrac{1}{\omega C}} {R} \right) \right)$$

$$P_a (\omega ) = \dfrac{2}{ \sqrt { 100^2 + \left( 50 \times 10^{-3} \; \omega - \dfrac{1}{470 \times 10^{-6}\; \omega } \right)^2}} \cos \left(\arctan \left( \dfrac{50 \times 10^{-3} \; \omega - \dfrac{1}{ 470 \times 10^{-6} \; \omega }} {100} \right) \right)$$

b)

$$\omega_r = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{ \sqrt{50 \times 10^{-3} \times 470 \times 10^{-6} }} \approx 206.28$$ 弧度/秒