# RC电路对阶跃电压的响应



## 带解决方案的问题

$v_i(t) - v_R(t) - v_C(t) = 0$       (I)

$v_R(t) = R \; i(t)$

$\displaystyle v_C (t) = \dfrac{1}{C} \int i dt$

$i (t) = C \dfrac{d v_C}{dt}$
$v_R (t) = R i (t) = R C \dfrac{v_C}{dt}$

$v_i (t) - R C \dfrac{v_C}{dt} - v_C (t) = 0$

$\mathscr{L} \left \{ v_i (t) - R C \dfrac{v_C}{dt} - v_C (t) \right \} = \mathscr{L}\{ 0 \}$

$\mathscr{L} \left\{ v_i (t) \right\} - R\;C \mathscr{L} \left \{ \dfrac{v_C}{dt} \right \} - \mathscr{L} \left\{ v_C (t) \right \} = 0$       (II)

$\mathscr{L} \left\{ \dfrac{v_C}{dt} \right \} = s V_C(s) - v_C(0)$

$V_i(s) - R C s V_C(s) - V_C(s) = 0$

$V_i(s) = \dfrac{V_0}{s}$

$R C s V_C(s) + V_C(s) = \dfrac{V_0}{s}$

$V_C(s) (R\;C\;s + 1) = \dfrac{V_0}{s}$

$V_C(s) = \dfrac{V_0}{s(R\;C s + 1)}$

$V_C(s) = \dfrac{V_0}{s(R\;C s + 1)} = \dfrac{V_0}{s} - \dfrac{R\;C V_0}{R\;C s + 1}$

$V_C(s) = V_0 \left(\dfrac{1}{s} - \dfrac{1}{s + \dfrac{1}{R\;C}} \right)$

$v_C(t) = V_0 \left( \mathscr{L^{-1}} \left\{ \dfrac{1}{s} \right\} - \mathscr{L^{-1}} \left\{ \dfrac{1}{s + \dfrac{1}{R\;C}} \right\} \right)$
$\mathscr{L^{-1}} \left\{ \dfrac{1}{s} \right\} = u(t)$

$\mathscr{L^{-1}} \left\{ \dfrac{1}{s + \dfrac{1}{R\;C}} \right\} = u(t) e^{-\frac{t}{R\;C}}$

$v_C(t) = V_0 (1 - e^{-\frac{t}{R\;C}} ) u(t)$

$v_R (t) = v_i - v_C = V_0 u(t) - V_0 u(t) (1 - e^{-\frac{t}{R\;C}} ) = V_0 e^{-\frac{t}{R\;C}} u(t)$

$i(t) = \dfrac{v_R}{R} = \dfrac{V_0}{R} e^{-\frac{t}{R\;C}} u(t)$

$R\;C = 200 \times 5 \times 10^{-3} = 1$ 秒(秒)
$v_C(t) = 10 (1 - e^{-t} ) u(t)$ 伏特
$v_R (t) = 10 e^{-t} u(t)$ 伏特
$i(t) = 0.05 e^{-t} u(t)$ 安培

1) 由于在 $t = 0$ 之前电容器未充电，因此电容器上的电压 $v_C(0)$ 为零，电容器在 $t = 0$ 时表现为短路。随着时间 $t$ 的增加，$v_C(t)$ 开始增加，这解释了电容器的充电过程
2) 电阻上的电压 $v_R(0)$ 等于电源电压 $10$ V，并随着时间 $t$ 的增加而开始下降。
3) 电流 $i(0)$ 处于最大值 $\dfrac{v_i(0) - v_C(0)}{R} = \dfrac{10 - 0}{200} = 10 / 200 = 0.05$ 安培，并随着时间 $t$ 的增加而减少。

$v_C(t)$ 几乎等于 $v_i(t)$，这意味着电容器已完全充电。电流 $i(t)$ 几乎为零，因为电容器表现为开路

$v_C(t) - v_R (t) = 0$       (I)

$v_R (t) = R \; i (t)$

$\displaystyle Q(t) = Q_0 - \int_0^{t} i(\tau) d\tau$

$v_C(t) = \dfrac{Q(t)}{C} = \dfrac{Q_0}{C} - \dfrac{ \displaystyle \int i dt \ }{C }$

$\dfrac{d v_C}{d t } = \dfrac{1}{C} \dfrac{d (Q_0 - \displaystyle \int i dt) }{d t }$

$\dfrac{d v_C}{d t } = \dfrac{1}{C} \dfrac{d Q_0}{dt} - \dfrac{1}{C} \dfrac{d(\displaystyle \int i dt) }{d t }$

$\dfrac{d v_C}{d t } = - \dfrac{1}{C} i$

$i(t) = - C \; \dfrac{v_C}{dt}$     注意 负号是因为电容器正在放电。

$v_R (t) = R \; i (t) = - R \; C \; \dfrac{d v_C}{dt}$

$v_C (t) + R \; C \; \dfrac{d v_C}{dt} = 0$

$\mathscr{L} \left \{ v_C (t) + R \; C \; \dfrac{d v_C}{dt} \right \} = \mathscr{L}\{ 0 \}$

$\mathscr{L} \left\{ v_C(t) \right \} + R\;C\;\mathscr{L} \left\{ \dfrac{d v_C}{dt} \right \} = 0$       (II)

$\mathscr{L} \left \{ \dfrac{d v_C}{dt} \right \} = s \; V_C(s) - v_C(0)$

$V_C(s) + R \; C \; ( s \; V_C(s) - V_0 ) = 0$

$V_C(s) \; (R\;C\;s + 1) = R \; C \; V_0$

$V_C(s) = \dfrac{R \; C \; V_0}{R \; C \; s + 1}$

$V_C(s) = \dfrac{ V_0}{s + \dfrac{1}{R\;C}}$
$v_C(t)$ 由拉普拉斯逆变换给出；因此：
$v_C(t) = V_0 \left( \mathscr{L^{-1}} \left\{ \dfrac{1}{s + \dfrac{1}{R\;C}} \right\} \right)$

$v_C(t) = V_0 \; e^{-\frac{t}{R\;C}}$

$v_R (t) = v_C = V_0 \; e^{-\frac{t}{R\;C}}$

$i(t) = \dfrac{v_R}{R} = \dfrac{V_0}{R} \; e^{-\frac{t}{R\;C}}$

$R\;C = 200 \times 5 \times 10^{-3} = 1$ 秒(秒)
$v_C(t) = 10 \; e^{-t}$ 伏特
$v_R (t) = 10 \; e^{-t}$ 伏特
$i(t) = 0.05 \; e^{-t}$ 安培

1) 由于在 $t = 0$ 之前电容器已充电至 $v_C(0) = 10$ 伏特，随着时间 $t$ 的增加，电压开始下降，这解释了电容器的放电过程
2) 在 $t = 0$ 时；$v_R (0) = v_C(0) = 10$。
3) 在 $t = 0$ 时，电流 $i(0)$ 处于最大值，给出为 $i(0) = \dfrac{v_R(0)}{R} = \dfrac{10}{200} = 10 / 200 = 0.05$ 安培，并随着时间 $t$ 的增加而减少。

## 附录 A

$\dfrac{V_0}{s(R\;C s + 1)} = \dfrac{A}{s} + \dfrac{B}{R\;C s + 1}$

$V_0 = A(R\;C\;s + 1) + B s$      (1)

$A = V_0$

$V_0 = V_0 \times (R\;C \times 1 + 1) + B \times 1$

$B = - R\;C V_0$

$\dfrac{V_0}{s(R\;C s + 1)} = \dfrac{V_0}{s} - \dfrac{R\;C V_0}{R\;C s + 1}$