Circuit Theory/2Source Excitement/Example45

From Wikibooks, open books for an open world
< Circuit Theory‎ | 2Source Excitement
Jump to: navigation, search
Problem L = 1H, C=1F, R=1Ω, find ir

Particular/Steady State solution[edit]

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

Homogeneous/Transient Solution[edit]

Loop equation:

V_s(t) = V_L(t) + V_{CR}(t)
V_s(t) = L {d i_t \over dt} + V_{RC}
V_{CR} = i_R*R
i_C = C {d V_{CR} \over dt}
i_t = i_R + i_C = \frac{V_{CR}}{R} + C {d V_{CR} \over dt}
V_s(t) = L ({d (\frac{V_{CR}}{R}) \over dt} + C {d^2 (V_{CR}) \over dt^2}) + V_{CR}

Differential equation that needs to be solved:

0 = L {d (\frac{V_{CR}}{R}) \over dt} + L C {d^2 (V_{CR}) \over dt^2} + V_{CR}

Guess:

V_{CR} = Ae^{st}

Substitute to check if possible:

0 = L {d (\frac{Ae^{-st}}{R}) \over dt} + L C {d^2 (Ae^{-st}) \over dt^2} + Ae^{-st}
0 = L (s\frac{Ae^{-st}}{R}) + L C (s^2 Ae^{-st}) + Ae^{-st}
0 = L (s\frac{A}{R}) + L C (s^2 A) + A
0 = \frac{L}{R}s + L Cs^2 + 1

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:

 s^2 + 2s + 1
 s_{1,2} = -1, -1

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

V_{CR} = Ae^{-t} + Bte^{-t}

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

(Ae^{-t} + Bte^{-t} - Be^{-t} - Be^{-t}) + 2(-Ae^{-t} - Bte^{-t} + Be^{-t}) + Ae^{-t} + Bte^{-t} = 0

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

V_{CR} = Ae^{-t} + Bte^{-t} + C_1

Without Initial Conditions .. Finding the Constants[edit]

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:

V_{CR}(0_+) = 0 = A + C_1

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.

V_{CR}(\infty) = C_1 = 5 \Rightarrow A = -5

This is the matlab code that computes the limit:

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

Only B is unknown now:

V_{CR} = -5e^{-t} + Bte^{-t} + 5
File:Example45D.png
Matlab plot of Ir which is Vcr multiplied by 2 which shows the capacitor slowly charging .. is critically damped charging .. most critically damped pictures are of discharing circuits .. that is why it looks different

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:

i_c = C {d V_{CR} \over dt} = 5e^{-t} + Be^{-t} - Bte^{-t}
i_c(0_+) = 5 + B = 0 \Rightarrow B = -5

Now VCR is:

V_{CR} = 5( 1 -e^{-t} - te^{-t})

Which means that ir is:

i_R = \frac{V_{CR}}{R} = 10( 1 -e^{-t} - te^{-t})

Without C_1 constant[edit]

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

V_L(0_+) = 5 = B (2 - 2)

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.