# Circuit Theory/2Source Excitement/Example45

Problem L = 1H, C=1F, R=1Ω, find ir

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:

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

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

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.