Solving for current in RLC circuit with Laplace and steady-state solution

Question

The following parallel RLC circuit is example 8.8 in Fundamentals of Electric Circuits by Charles Alexander.

In the example in this text, the current through the inductor i(t) is found to be:

$$i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978})$$

This solution is found by directly solving the second order differential equation. I want to solve this same circuit using Laplace transforms. Following the methods in the textbook, I have performed a Laplace transform on this circuit:

^{simulate this circuit – Schematic created using CircuitLab}

Please note that even though 4 A current supply is always I, I have converted it to from 4 A to 4/s; I hope that is correct. Also, the initial current through the inductor is 4 A, leading to another 4/s current supply in the opposite direction.

In this circuit, the current through the inductor is:

$$ I(s) = \frac{0.75 s}{s(s^2 + 12.5 s + 6.25)}$$

The inverse Laplace transform of I(s) is:

$$i(t) = 0.0655(e^{-0.5218t} - e^{-11.978})$$

In this second solution, the "4" is missing. I have a two questions:

Should Laplace only give the transient solution?

Even though the current supply on the left does not change at t=0, why must it change to 4/s in the Laplace domain?

Answer 1

I wish you would have disclosed your process. It would have been appreciated.

Grounding the bottom node, find this KCL for the switched node:

$$\begin{align*} \frac{v_t}{R_1}+\frac{v_t}{R_2}+C_1\frac{\text{d}\,v_t}{\text{d}t}+\frac1{L_1}\int v_t\:\text{d}t&= 0\:\text{A} \\\\ \frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t&= 0\:\text{A} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t+\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\mathscr{L}\left\{\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\frac{\text{d}}{\text{d}t}v_t\right\}+\mathscr{L}\left\{\frac1{L_1\,C_1}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \mathscr{L}\left\{\frac{\text{d}^2}{\text{d}t^2}v_t\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\mathscr{L}\left\{\frac{\text{d}}{\text{d}t}v_t\right\}+\frac1{L_1\,C_1}\mathscr{L}\left\{\vphantom{\frac{\text{d}}{\text{d}t}}v_t\right\}&= \mathscr{L}\left\{0\:\text{A}\right\} \\\\ \left\{\vphantom{\frac{\text{d}}{\text{d}t}}s^2V_s-sv_{_0}-v_{_0}^{'}\right\}+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\left\{\vphantom{\frac{\text{d}}{\text{d}t}}sV_s-v_{_0}\right\}+\frac1{L_1\,C_1}V_s&= \mathscr{L}\left\{0\:\text{A}\right\} \end{align*}$$

Or,

$$\begin{align*} \left[s^2+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)s+\frac1{L_1\,C_1}\right]V_s&= \left[s+\frac1{C_1}\left(\frac1{R_1}+\frac1{R_2}\right)\right]v_{_0}+v_{_0}^{'} \end{align*}$$

Since \$v_{_0}=15\:\text{V}\$ and the current supplied by \$C_1\$ at \$t=0\$ must be \$-1.5\:\text{A}\$, it's clear that \$\frac{\text{d}}{\text{d}t}v_{t=0}=\frac{-1.5\:\text{A}}{8\:\text{mF}}=-187.5\:\frac{\text{V}}{\text{s}}\$. The above then becomes:

$$\begin{align*} \left[s^2+12.5s+6.25\right]V_s&= 15s \end{align*}$$

Now for your mistake. I suspect that you elected to believe that \$V_s=s L_1 I_s\$ and found that:

$$\begin{align*} s\left[s^2+12.5s+6.25\right]I_s&= 0.75s \end{align*}$$

However, that's not correct.

Instead \$V_s=\mathscr{L}\left\{v_t\right\}=L_1 \mathscr{L}\left\{i_t^{'}\right\}=L_1\left[sI_s-i_{_0}\right]\$.

So, as \$L_1=20\:\text{H}\$:

$$\begin{align*}\require{cancel} \left[s^2+12.5s+6.25\right]L_1\left[sI_s-i_{_0}\right]&= 15s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s-i_{_0}\right]&= 0.75s \\\\ \left[s^2+12.5s+6.25\right]\left[sI_s\right]&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ s\left[s^2+12.5s+6.25\right]I_s&= 0.75s+\left[s^2+12.5s+6.25 \right]i_{_0} \\\\ I_s&=\frac{0.75\cancel{s}}{\cancel{s}\left[s^2+12.5s+6.25\right]}+\frac{\cancel{\left[s^2+12.5s+6.25 \right]}4}{s\cancel{\left[s^2+12.5s+6.25\right]}} \\\\ &=\frac{0.75}{s^2+12.5s+6.25}+\frac4{s} \end{align*}$$

So, your Laplace equation should be \$I_s=\frac4{s}+\frac{0.75}{s^2+12.5\, s+6.25}\$.

If you had solved this in the time domain, you'd have been using annihilation and finding a need for hyperbolic sine (hyperbolic because the system is over-damped.)

$$i_t=4.0 + 0.130930734141595\cdot\exp\left(-6.25\,t\right)\cdot\sinh\left(5.7282196186948\,t\right)$$

The way you wrote it is also correct as the hyperbolic sine has a built-in \$\frac12\$ factor and expanding it would change 0.130930734141595 to 0.0654653670707975 (and provide your difference pair of exponentials.)

Answer 2

It seems that it should be this ...

Final current into L1 = 4 + 30/20 = 5.5 A ?

Or this ?