Solving nonsteady circuits.

[piotrekn95]

Hello,

I am preparing for my circuit theory. I am trying now to solve a basic circuit like this :



I know that I could use Laplace theory to do it (with s symbol).
But unfortunately this circuit is too tricky for me to begin. I see that if we have DC input, the coil is like a wire, and there is no way through capacitor before switching the switch.



As far I circled only the values of voltage and current at the beginning (- before switching, +after switching), but can you help me what next step should i do to find the il(t).

Thank you

 

[CataM]

You have found out the initial conditions of the circuit. You can also find out the steady state values after the switch has been turned on which are 3 A and 30 V.
To solve circuits with inductors and capacitors you have to write the differential equations. You have to redraw the circuit at t=0+.

You may start with this:

30 V = R1·iL + (L·diL/dt) (very simple 1st order linear and with constant coefficients differential equation; iL = current through the inductor)

Likewise for the voltage of the capacitor, exactly the same.

---------------

The method showed on top is the mathematical approach and in my opinion is the way to go because the method is remembered always meanwhile formulas are easy forgattables.

However, you can memorise for the exam (and if the teacher allows you) the general solution of an Ordinary Differential Equation with constant coefficients (this case) which saves you time and is as follows:

x(t)=X_f + (X₀-X_f)·e^-(t-t0)/tau

X_f=final value i.e. steady state value
X₀=initial condition
t0=time at which the initial condition is found out (in this case is t0 = 0 s)
tau=time constant

In conclusion:

For the inductor is as follows: iL(t)=3 A + (1 A - 3 A) · e^-(t-0)/L/R1 = 3-2·e^{bla bla}
Capacitor is as follows: vC(t)=30 V + (10 V - 30 V)·e^{bla bla}