Creating a RL/RC/RLC Circuit model

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To create a model for an RL/RC/RLC circuit using differential equations, it's essential to understand that the total voltage change around the circuit must equal zero. The voltage drops across resistors, inductors, and capacitors can be expressed as Ri, L di/dt, and Q/C, respectively. Current is defined as the rate of change of charge, leading to relationships that involve derivatives of charge over time. By combining these voltage drops and equating them to the supplied voltage, one can derive a second-degree differential equation for charge (Q). Understanding these concepts is crucial for applying differential equations to circuit analysis effectively.
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Hello! Would someone here guide me in creating a model for a RL/RC/RLC circuit? I have started studying this topic in advanced but I don't quite understand the concept in DE.
I have watched videos from youtube.com and they solve problems using EE such as V=IR. And yes, I am able to answer problems too using theoretical formulas from my EE class. What I want to know is how to apply DE in this topic. I really want to learn this soon because maybe next week, we are going to start on this topic.
Thanks!:-p
 
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Your question is really too vague to give a complete answer. Exactly how you model such a circuit as a differential equation depends on exactly what the circuit is like.

The basic ideas are
1) The total voltage change around a circuit is 0.
2) The voltage drop over a resistor is Ri where i is the current and R the resistance.
3) The voltage drop through a coil is L di/dt where i is the current and L the inductance.
4) The voltatge drop through a capacitor is Q/C where Q is the charge and C is the capacitance (you will have to ask an electrical engineer why C is divided and not multiplied).

Since current is the rate of change of charge, i= dQ/dt and so di/dt= d^2Q/dt^2 and the voltage drop through a resistor is R(dQ/dt) and the voltage change through a coil is L(d^2Q/dt^2).

Since those are all "drops" (negative change in voltage), there must be voltage supplied to the circuit for the total change to be 0. Adding those for a circuit that has all three kinds of objects and setting it equal to the voltage supplied gives you a second degree differential equation for Q.

(It is not a good idea to "bump" a thread after only 8 minutes. That can get you banned!)
 
Thanks! I get it now! Thank you very much physicsforum and HallsofIvy! :biggrin:
 

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