Circuit Theory/2Source Excitement/Example45

Problem L = 1H, C=1F, R=1Ω, find i_r

Particular/Steady State solution

Inductor short, cap open, V_s = 5 μ(t),find i_r

Homogeneous/Transient Solution

Loop equation:

$V_s(t)=V_L(t)+V_{CR}(t)$

$V_s(t)=L\frac{d{i}_t}{dt}+V_{RC}$

$V_{CR}=i_R*R$

$i_C=C\frac{d{V}_{CR}}{dt}$

$i_t=i_R+i_C=\frac{V_{CR}}{R}+C\frac{d{V}_{CR}}{dt}$

$V_s(t)=L(\frac{d(\frac{V_{CR}}{R})}{dt}+C\frac{d^2(V_{CR})}{d{t}^2})+V_{CR}$

Differential equation that needs to be solved:

$0=L\frac{d(\frac{V_{CR}}{R})}{dt}+LC\frac{d^2(V_{CR})}{d{t}^2}+V_{CR}$

Guess:

$V_{CR}=A{e}^{st}$

Substitute to check if possible:

$0=L\frac{d(\frac{A{e}^{-st}}{R})}{dt}+LC\frac{d^2(A{e}^{-st})}{d{t}^2}+A{e}^{-st}$

$0=L(s\frac{A{e}^{-st}}{R})+LC(s^{2}A{e}^{-st})+A{e}^{-st}$

$0=L(s\frac{A}{R})+LC(s^{2}A)+A$

$0=\frac{L}{R}s+LC{s}^2+1$

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:

$V_{CR}=A{e}^{-t}+Bt{e}^{-t}$

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

$(A{e}^{-t}+Bt{e}^{-t}-B{e}^{-t}-B{e}^{-t})+2(-A{e}^{-t}-Bt{e}^{-t}+B{e}^{-t})+A{e}^{-t}+Bt{e}^{-t}=0$

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

$V_{CR}=A{e}^{-t}+Bt{e}^{-t}+C_1$

Without Initial Conditions .. Finding the Constants

Have initial conditions: V_CR(0₊) = 0 since initially cap is a short and impedance times the derivative of the inductor current i_t(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.

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

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:

$V_{CR}=-5e^{-t}+Bt{e}^{-t}+5$

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:

$i_c=C\frac{d{V}_{CR}}{dt}=5e^{-t}+B{e}^{-t}-Bt{e}^{-t}$

$i_c(0_{+})=5+B=0\Rightarrow B=-5$

Now V_CR is:

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

Which means that i_r is:

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

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. Template:BookCat