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:

$V_{s}(t)=V_{L}(t)+V_{CR}(t)$

$V_{s}(t)=L{di_{t} \over dt}+V_{RC}$

$V_{CR}=i_{R}*R$

$i_{C}=C{dV_{CR} \over dt}$

$i_{t}=i_{R}+i_{C}={\frac {V_{CR}}{R}}+C{dV_{CR} \over dt}$

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

Differential equation that needs to be solved:

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

Guess:

$V_{CR}=Ae^{st}$

Substitute to check if possible:

$0=L{d({\frac {Ae^{-st}}{R}}) \over dt}+LC{d^{2}(Ae^{-st}) \over dt^{2}}+Ae^{-st}$

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

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

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

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 V_cr is:

$V_{CR}=Ae^{-t}+Bte^{-t}+C_{1}$

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.

$V_{CR}(\infty )=C_{1}=5\Rightarrow A=-5$

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:

$i_{c}=C{dV_{CR} \over dt}=5e^{-t}+Be^{-t}-Bte^{-t}$

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

Now V_CR is:

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

Which means that i_r 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.