The formulas developed in the series RLC circuit response to a step voltage are presented here as they are used in the calculator.
In the formulas below, \( \alpha = \dfrac{R}{2 L} \)
When a voltage step function of the form \( V_0 u(t) \) we have three possible cases to consider:
Case 1: The circuit is underdamped when \( \dfrac{1}{L C} \gt \left(\dfrac{R}{2 L}\right)^2 \)
Let \( \omega = \sqrt {\dfrac{1}{L C} - \left(\dfrac{R}{2 L}\right)^2} \)
The current and voltages are given by
\( i(t) = \dfrac{V_0}{\omega L} \; \sin (\omega t) \; e^{-\alpha t} \)
\( v_R(t) = R \; i(t) = V_0 \dfrac{2 \alpha}{\omega } \sin (\omega t) e^{-\alpha t} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = V_0 ( \cos (\omega t)- \dfrac{\alpha}{\omega} \sin (\omega t) ) e^{-\alpha t} \)
\( v_C(t) = V_0 - V_0 \left\{ \cos (\omega t) + \dfrac{\alpha}{\omega} \sin (\omega t) \right\} e^{-\alpha t} \)
Case 2: The circuit is overdamped when \( \dfrac{1}{L C} \lt \left(\dfrac{R}{2 L}\right)^2 \)
Let \( \beta = \sqrt { \left(\dfrac{R}{2 L}\right)^2 - \dfrac{1}{L C} } \) and rewrite \( I(s) \) as
The current and voltages are given by
\( i(t) = \dfrac{V_0}{\beta L} \; \sinh (\beta t) \; e^{-\alpha t} \)
\( \quad \quad = \dfrac{V_0}{2\beta L} \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( v_R(t) = V_0 \dfrac{\alpha}{\beta } \; \left\{ e^{ (\beta - \alpha) t} - e^{ ( - \beta - \alpha) t} \right\} \)
\( v_L(t) = L \; \dfrac{d i}{dt} = \dfrac{V_0}{2\beta} \left\{ (\beta - \alpha) e^{ (\beta - \alpha) t} + (\beta + \alpha) e^{ ( - \beta - \alpha) t} \right\} \)
\( v_C(t) = V_0 - V_0 \left\{ \dfrac{\beta + \alpha}{2 \beta} e^{(\beta - \alpha) t} + \dfrac{\beta - \alpha}{2 \beta} e^{(-\beta - \alpha) t} \right\} \)
Case 3: The circuit is critically damped \( \dfrac{1}{L C} = \left(\dfrac{R}{2 L}\right)^2 \)
The current and voltages are given by
\( i(t) = \dfrac{V_0}{ L} \; t \; e^{-\alpha t} \)
\( v_R(t) = 2 V_0 \alpha \; t \; e^{-\alpha t} \)
\( v_L(t) = V_0 e^{- \alpha t} \left( 1 - \alpha t \right) \)
\( v_C(t) = V_0 - V_0(1+\alpha t)e^{-\alpha t} \)