Second Order Circuits: Difference between revisions

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<math>iR + L \frac{di}{dt} + V_C = 0</math>
<math>iR + L \frac{di}{dt} + V_C = 0</math>


Next, we can use <math>i = C \frac{dV_C}{dt}</math> to get
Next, we can use <math>i = C \frac{dV_C}{dt}</math> and substitution to get


<math>RC \frac{dV_C}{dt} + L \frac{d}{dt} \frac{C V_C} {dt} V_C = 0</math>
<math>RC \frac{dV_C}{dt} + L \frac{d}{dt} \frac{C V_C} {dt} V_C = 0</math>
Changing the order and moving the constants,


<math>LC \frac{d^2 V}{dt^2} + RC \frac{dV_C}{dt} + V_C = 0</math>
<math>LC \frac{d^2 V}{dt^2} + RC \frac{dV_C}{dt} + V_C = 0</math>
Moving constants away from the first term to get a ''second-order differential equation,''


<math>\frac{d^2V_C}{dt^2} + \frac{R}{L} \frac{dV_C}{dt} + \frac{1}{LC} V_C = 0</math>
<math>\frac{d^2V_C}{dt^2} + \frac{R}{L} \frac{dV_C}{dt} + \frac{1}{LC} V_C = 0</math>
The above is a ''second-order differential equation!'' We have ways to solve that.

Revision as of 03:46, 1 March 2024

Second order circuits are circuits that have two energy storage

elements, resultingin second-order differential equations. The circuits have to be cast in terms of irreducible elements (i.e. combine L's and C's into one when possible).

An unforced RLC circuit

Consider an un-forced RLC circuit. We want to find .

First, we can use KVL and KCL

Next, we can use and substitution to get

Changing the order and moving the constants,

Moving constants away from the first term to get a second-order differential equation,