Circuit Theory/2Source Excitement/Example46

RLC problem for circuit theory wikibook

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

Steady State Particular Solution

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

Transient Particular Solution

Start writing a node equation:

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

Substitute voltage terminal relationships:

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

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

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

Substitute to get I_s in terms of i_o:

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

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

Substituting numbers from the problem:

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

Time constant

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

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

$I_{s}=s^{2}Ae^{-st}+2sAe^{-st}+2Ae^{-st}$

Does this equal zero?

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

$s^{2}+2s+2=0$

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

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

$\sigma =-1$

$\omega =1$

So the next guess is:

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

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.

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

So now the expression for i_o is:

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

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

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

Which means that:

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

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

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

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

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

So:

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

Thus i_o is:

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

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