# Circuit Theory/2Source Excitement/Example45

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

## Homogeneous/Transient Solution

Loop equation:

${\displaystyle V_{s}(t)=V_{L}(t)+V_{CR}(t)}$
${\displaystyle V_{s}(t)=L{di_{t} \over dt}+V_{RC}}$
${\displaystyle V_{CR}=i_{R}*R}$
${\displaystyle i_{C}=C{dV_{CR} \over dt}}$
${\displaystyle i_{t}=i_{R}+i_{C}={\frac {V_{CR}}{R}}+C{dV_{CR} \over dt}}$
${\displaystyle V_{s}(t)=L({d({\frac {V_{CR}}{R}}) \over dt}+C{d^{2}(V_{CR}) \over dt^{2}})+V_{CR}}$

Differential equation that needs to be solved:

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

Guess:

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

Substitute to check if possible:

${\displaystyle 0=L{d({\frac {Ae^{-st}}{R}}) \over dt}+LC{d^{2}(Ae^{-st}) \over dt^{2}}+Ae^{-st}}$
${\displaystyle 0=L(s{\frac {Ae^{-st}}{R}})+LC(s^{2}Ae^{-st})+Ae^{-st}}$
${\displaystyle 0=L(s{\frac {A}{R}})+LC(s^{2}A)+A}$
${\displaystyle 0={\frac {L}{R}}s+LCs^{2}+1}$

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:

${\displaystyle s^{2}+2s+1}$
${\displaystyle s_{1,2}=-1,-1}$

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

${\displaystyle V_{CR}=Ae^{-t}+Bte^{-t}}$

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

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

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

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

## 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:

${\displaystyle V_{CR}(0_{+})=0=A+C_{1}}$

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.

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

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:

${\displaystyle V_{CR}=-5e^{-t}+Bte^{-t}+5}$

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:

${\displaystyle i_{c}=C{dV_{CR} \over dt}=5e^{-t}+Be^{-t}-Bte^{-t}}$
${\displaystyle i_{c}(0_{+})=5+B=0\Rightarrow B=-5}$

Now VCR is:

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

Which means that ir is:

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

## Without C_1 constant

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

${\displaystyle V_{L}(0_{+})=5=B(2-2)}$

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.