Circuit Theory/2Source Excitement/Example45

Particular/Steady State solution
Inductor short, cap open, Vs = 5 &mu;(t),find ir

Homogeneous/Transient Solution
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
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:
 * $$V_{CR} = -5e^{-t} + Bte^{-t} + 5$$

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
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.