Circuit Theory/2Source Excitement/Example45

From Wikibooks, open books for an open world
Jump to navigation Jump to search
Problem L = 1H, C=1F, R=1Ω, find ir

Particular/Steady State solution[edit | edit source]

Inductor short, cap open, Vs = 5 μ(t),find ir

Homogeneous/Transient Solution[edit | edit source]

Loop equation:

Differential equation that needs to be solved:


Substitute to check if possible:

So the answer is going to be second order, thus guess was wrong, but can guess more accurately now by computing roots of the above second order equation:

Both roots are negative and equal, so the new guess is:

Checking again by plugging into s2 + 2s + 1 = 0:

Yes it equals zero now! So can go on. Have to add a constant to the differential equation solution so Vcr is:

Without Initial Conditions .. Finding the Constants[edit | edit source]

Have initial conditions: VCR(0+) = 0 since initially cap is a short and impedance times the derivative of the inductor current it(0+) = 5. Turning this into an equation:

The final voltage across the parallel RC combination is going to be 5 volts (after a very long time) because the capacitor opens and the inductor shorts.

This is the matlab code that computes the limit:

syms A B C1 t
f = A*exp(-t) + B*t*exp(-t) + C1;

Only B is unknown now:

The initial voltage across the inductor is going to be 5 volts. But this does not lead to the value of B. Another initial condition is that the initial current through the capacitor (even though it is initially a short) is zero because the inductor is initially an open. This leads to B:

Now VCR is:

Which means that ir is:

Without C_1 constant[edit | edit source]

Trying to do this problem without the C_1 constant ends in something like this:

Which has no solution. Or it can lead to 5=5 where the constant disappears from the equation without finding a number for it. Or it can lead to A or B equaling infinity. Any of these non-answers means a mistake was made somewhere.