- **Classical Physics**
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- - **Linear circuit problems**
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linear circuit problemsIn solving linear circuit problems using complex currents, charges etc. I have stumbled upon something I never really understand. You are basically solving 1st and 2nd order differential equations by algebraic means, right? Well at what point do you apply initial conditions to your current and charge, Q(0), dQ/dt (0)?
I might have missed out on an essential points, because it actually seems that you never apply boundary conditions. |

Re: linear circuit problemsYou could discover where the boundary conditions come in (and they don't have to be at t=0) by solving the same DEs the usual way :)
What you are doing is exploiting that you know the solutions to the DEs with the appropriate boundary conditions already because you know the relationship between PD and dQ/dt. |

Re: linear circuit problemsuhh. what is PD? :(
Let me just clarify exactly the method I have learned and you can point at which step I use boundary conditions: 1) Say we have an RCL with I(0)=0, Q(0)=a driven by a time varying potential U(t) = εcos(ωt) 2) We write (where I is now complex I = I0exp(-iωt) and β is a phase) εexp(-iωt) = RI0exp(-iωt) - iωLβI0exp(-iωt) - 1/(-iωC) I0exp(-iωt) 3) You can cancel out the exponential function and find the phase which can be found entirely in terms of the physical quantities L,C,R. The same goes for I0, the amplitude of the current. 4) You can now put it all together and take the real part of the current. It will be a sinusoidal function with determined amplitude and phase - which is one specific solution and is not a general solution. WHERE did I apply the boundary conditions? |

Re: linear circuit problemsPD=potential difference.
Quote:
You did explicitly assume I(0)=0 and Q(0)=a (Q(0) where?). When you draw the voltage source in the diagram, you specify a phase direction. When you draw the PD arrows on the diagram - you use that phase direction to decide if the potential is gained or dropped. Those contribute to the boundary conditions. Since the current will be sinusoidal, it actually doesn't matter what the initial phase is - so we have picked one that makes the math easy. You'll see it clearly when you do the calculus. |

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