Second Order Circuits
Second order circuits are circuits that have two energy storage elements, resulting in second-order differential equations.
One application of second order circuits is in timing computers. As we will see, an RLC circuit can generate a sinusoidal wave.
There are primarily two types of second order circuits:
- Parallel RLC circuits
- Series RLC circuits
This page will analyze them and derive some useful equations.
Series RLC Circuits
Natural Response

Consider an un-forced RLC circuit. We want to find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_C} .
First, we can use KVL and KCL
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_R + V_L + V_C = 0}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle iR + L \frac{di}{dt} + V_C = 0}
Next, we can use Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i = C \frac{dV_C}{dt}} and substitution to get
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle RC \frac{dV_C}{dt} + L \frac{d}{dt} \frac{C V_C} {dt} V_C = 0}
Changing the order and moving the constants,
Moving constants away from the first term to get a second-order differential equation,
Parallel RLC Circuits
Natural Response

By KCL,
By differentiating once with respect to and rearranging some constants,
we get a homogeneous second-order differential equation, which has a standard solution that I will not go into detail. Briefly, it is solved by assuming since derivatives of must take the same form to cancel out to zero.
By applying the standard solution, we have
Which simplifies to
This is the characteristic equation of the differential equation, as the root of the quadratic determines properties of
where
and
It can be pretty easily proven that the sum of the two roots is also a solution
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v = A_1 e^{s_1 t} + A_2 e^{s_2 t} }
