Circuit Theory/2Source Excitement/Example46

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RLC problem for circuit theory wikibook

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

Steady State Particular Solution[edit]

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[edit]

Start writing a node equation:

 I_s = i_c + i_r + i_o

Substitute voltage terminal relationships:

 I_s = C {d V \over dt} + \frac{V}{R} + i_o

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

 V = L {d i_o \over dt} + R i_o

Substitute to get Is in terms of io:

 I_s = C {d (L {d i_o \over dt} + R i_o) \over dt} + \frac{L {d i_o \over dt} + R i_o}{R} + i_o
 I_s = CL {d^2 i_o \over dt^2} + CR {d i_o \over dt} + \frac{L}{R} {d i_o \over dt} + i_o + i_o

Substituting numbers from the problem:

 I_s = {d^2 i_o \over dt^2} + 2 {d i_o \over dt} + 2 i_o

Time constant[edit]

Guess: i_o(t) = Ae^{st} Substituting:

 I_s = {d^2 Ae^{-st} \over dt^2} + 2 {d Ae^{-st} \over dt} + 2 Ae^{-st}
 I_s =  s^2Ae^{-st} + 2sAe^{-st} + 2Ae^{-st}

Does this equal zero?

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

No. Rats. Need to evaulate 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[edit]

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 io 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 io(0+) = 0:

 i_o(o_+) = \frac{1}{2}+ A_1 = 0

Which means that:

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:

V_L = L {d i_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


A_2 = A_1 = -\frac{1}{2}

Thus io 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 io for a complicated source using the convolution integral.