# Circuit Theory/2Source Excitement/Example46

RLC problem for circuit theory wikibook

Find the time domain expression for io given that Is = 1μ(t) amp.

## Contents

After a long time, the cap opens and inductor shorts putting the two resistors in parallel and splitting the current making io 1/2 amp.

## Transient Particular Solution

Start writing a node equation:

${\displaystyle I_{s}=i_{c}+i_{r}+i_{o}}$

Substitute voltage terminal relationships:

${\displaystyle I_{s}=C{dV \over dt}+{\frac {V}{R}}+i_{o}}$

Find V in terms of io through the L R branch:

${\displaystyle V=L{di_{o} \over dt}+Ri_{o}}$

Substitute to get Is in terms of io:

${\displaystyle I_{s}=C{d(L{di_{o} \over dt}+Ri_{o}) \over dt}+{\frac {L{di_{o} \over dt}+Ri_{o}}{R}}+i_{o}}$
${\displaystyle I_{s}=CL{d^{2}i_{o} \over dt^{2}}+CR{di_{o} \over dt}+{\frac {L}{R}}{di_{o} \over dt}+i_{o}+i_{o}}$

Substituting numbers from the problem:

${\displaystyle I_{s}={d^{2}i_{o} \over dt^{2}}+2{di_{o} \over dt}+2i_{o}}$

## Time constant

Guess: ${\displaystyle i_{o}(t)=Ae^{st}}$ Substituting:

${\displaystyle I_{s}={d^{2}Ae^{-st} \over dt^{2}}+2{dAe^{-st} \over dt}+2Ae^{-st}}$
${\displaystyle I_{s}=s^{2}Ae^{-st}+2sAe^{-st}+2Ae^{-st}}$

Does this equal zero?

${\displaystyle s^{2}Ae^{st}-2sAe^{st}+2Ae^{st}=0}$
${\displaystyle s^{2}+2s+2=0}$

No. Rats. Need to evaulate the above quadratic in order to guess another solution.

${\displaystyle s_{1,2}=-1-j,-1+j=\sigma \pm j\omega }$
${\displaystyle \sigma =-1}$
${\displaystyle \omega =1}$

So the next guess is:

${\displaystyle i_{o}=e^{-t}(A_{1}\cos t+A_{2}\sin t)+C_{1}}$

## Finding the Constants

After a very long time, the capacitor is going to open and the inductor is going to short. This leaves two equal resistors in parallel that are going to split the current in half.

${\displaystyle i_{o}(\infty )={\frac {1}{2}}=C_{1}}$

So now the expression for io is:

${\displaystyle i_{o}=e^{-t}(A_{1}\cos t+A_{2}\sin t)+{\frac {1}{2}}}$

Initially the current through the conductor is 0, so io(0+) = 0:

${\displaystyle i_{o}(o_{+})={\frac {1}{2}}+A_{1}=0}$

Which means that:

${\displaystyle A_{1}=-{\frac {1}{2}}}$

The other initial condition affecting io is the voltage across the inductor .. which is zero. We can find an expression for VL:

${\displaystyle V_{L}=L{di_{0}(t) \over dt}=L(-e^{-t}(A_{1}\cos t+A_{2}\sin t)+e^{-t}(-A_{1}\sin t+A_{2}\cos t))}$

Setting all this equal to 0 at t=0 yields:

${\displaystyle 0=-A_{1}+A_{2}}$

So:

${\displaystyle A_{2}=A_{1}=-{\frac {1}{2}}}$

Thus io is:

${\displaystyle i_{o}={\frac {1}{2}}+e^{-t}(-{\frac {1}{2}}\cos t-{\frac {1}{2}}\sin t)={\frac {1}{2}}(1-e^{-t}(\cos t+\sin t))}$

This solution is used to find io for a complicated source using the convolution integral.