Solving Linear Circuit Problems: A Guide to Using Complex Currents and Charges"

[aaaa202]

In 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.

 

[Simon Bridge]

You 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.

 

[aaaa202]

uhh. 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?

 

[Simon Bridge]

PD=potential difference.

WHERE did I apply the boundary conditions?

... write out the general solution (from the DE) for the kind of quantity you have to solve when you do linear circuit analysis, then compare with the result you get when you use that analysis, and you'll be able to see what the boundary conditions were.

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.

 

[PJC01]

I can assure you that solving linear circuit problems using complex currents and charges is a valid and effective approach. However, I understand your confusion about applying initial conditions in this process.

Firstly, let me clarify that solving 1st and 2nd order differential equations by algebraic means is a fundamental technique in solving linear circuit problems. This allows us to simplify the equations and solve for the unknown variables.

Now, to address your question about initial conditions, it is important to note that in linear circuits, the initial conditions for current and charge are typically assumed to be zero. This is because linear circuits are often analyzed in a steady-state condition, where the circuit has reached a stable and unchanging state.

However, there are cases where initial conditions may need to be considered, such as when analyzing transients in a circuit or when the circuit is not in steady-state. In these cases, the initial conditions can be applied to the equations and solved for in the same way as any other variable.

In summary, initial conditions may not always be explicitly stated in linear circuit problems, but they can still be applied and solved for if needed. I hope this helps clarify your understanding of solving linear circuit problems using complex currents and charges.