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Please consider the following circuit:

enter image description here

At t>0 this circuit will be transformed to source-free parallel RLC-circuit, where capacitor voltage is Vc(0+) = 0 V and inductor current is Il(0+) = 4. Now is the time to find the response of the circuit.

Here is the context: I use "Fundamentals of electric circuits" of Charles K. Alexander and Matthew N.O. Sadiku. All the example problems follow the standard procedure: 1) define damping factor and resonant frequency; 2) define the type of response (overdamped, critically damped or underdamped); 3) choose the appropriate response equation; 4)...

I have calculated damping factor and resonant frequency, resolved corresponding characteristic equation and got these natural frequencies: s1 = -1, s2 = -6. This is the case of the overdamped response and I am ready to write the response equation:

v(t) = A1*exp(-t) + A2*exp(-6t); (1)

i(t) = A1*exp(-t) + A2*exp(-6t); (2)

However, I'm perplexed: which variable should I choose: voltage or current? Since the initial conditions (v(0+), i(0+), dv(0+)/dt, di(0+)/dt) are not the same, A1 and A2 will have different values for the equations 1 and 2. For this particular case if the voltage across the capacitor is used as a variable, then A1 = -33.6 and A2 = 33.6 which is incorrect (LTSpice confirm it). If I use a current through the inductor, then A1 = -4.8 and A2 = 0.8 (correct answer).

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  • \$\begingroup\$ Why try to solve a complicated differential equation when you can simply write the Laplace transform? \$\endgroup\$ – Bart Aug 22 '19 at 9:21
  • \$\begingroup\$ @ Bart, excellent question! Also, as authors say, "Problems in this chapter can also be solved by using Laplace transforms, which are covered in Chapters 15 and 16". I'm currently on the Chapter 8 and still too young to deal with Laplace transforms. \$\endgroup\$ – tenghiz Aug 22 '19 at 9:25
  • \$\begingroup\$ @Bart, by the way, if you mention Laplace transform, is it the common way in industry to deal with RLC-networks? Have I chosen a wrong textbook? \$\endgroup\$ – tenghiz Aug 22 '19 at 9:36
  • \$\begingroup\$ Laplace transforms are easier, because you just write the complex impedance for each component, and then use Kirchoff's laws to solve the circuit. After that, you apply the inverse transform, which usually means looking them up rather than solving the inverse transform integral. I'm not sure what the industry prefers, but I guess whereever money is involved, you want to be as cheap as possible. \$\endgroup\$ – Bart Aug 22 '19 at 10:37
  • \$\begingroup\$ By the way, solving the circuit using complex impedances is actually the same as directly writing the characteristic equation, skipping the D.E. \$\endgroup\$ – Bart Aug 22 '19 at 10:40
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I would like to thank @Jonk for the hint. I have discovered that I was very close to the solution, but missed only one step.

Let's start. Here is a parallel RLC-network and we have to find the equation for the waveform (voltage and current). As I have mentioned in the question, whether it is a voltage waveform or a current one, damping factor (alpha) and resonant frequency (omega) always be the same and they always lead to 3 different cases:

alpha > omega; overdamped response
alpha = omega; critically damped response
alpha < omega; underdamped response

Let's calculate these values:

alpha = 1/2RC = 3.5 Hz;
omega = 1/(LC)^0.5 = (6)^0.5.

Here we can see that it is a overdamped response case. Let's calculate natural frequencies:

s = -alpha +/- (alpha^2 - omega^2)^0.5 = -1; -6

In this case the general solution has the following form:

x(t) = A1*exp(-1t) + A2*exp(-6t).

Now we have to decide what is X - voltage or current? Let it be voltage:

v(t) = A1*exp(-1t) + A2*exp(-6t); general solution for the voltage waveform

Let's define A1 and A2. We have to compose a system of two independent equations for the volatge and the voltage derivative at t=0. Initial conditions can be obtained from the circuit: when t<0, the capacitor is shunted by the inductor, so v(0-) = v(0+) = 0. At t>0, there is no voltage source in the circuit, thus final voltage = 0. 

v(0) = A1*exp(-1*0) + A2*exp(-6t*0) = A1 + A2;

Let's find out the voltage derivative. Lazy way: just open the textbook and pick the formula:

dv/dt = - (v(0) + Ri(0))/RC = -168;

Not lazy way (it will give the same result). There are 3 currents in the parallel source-free RLC network from the top node downwards to the ground:

0 A = CdV/dt + iL(0) + v(0+)/R = CdV/dt + 4 + 0/R;
CdV/dt = -4;
dV/dt = -4 * 42 = -168.

Let's differentiate the general solution equation:

dv/dt = d(A1*exp(-1t) + A2*exp(-6t))/dt = -A1*exp(-1t) - 6*A2*exp(-6t));
dv(0)/dt = -A1 - 6*A2;

Now it is time to compose a system of two equations:

0 = A1 + A2;
-168 = -A1 - 6*A2;
--------------------> A2 = 33.6, A1 = -33.6.

Also, the general solution for the voltage waveform is:

v(t) = 33.6*exp(-6t) - 33.6*exp(-t)).

Now let's remember the advice of @Jonk. This is the parallel RLC-circuit and the volatge is the same across all the branches. Thus, inductor has the same voltage. The current across the inductor is the integral of the voltage:

i(t) = (1/L)* ∫ v(t)dt = (33.6/7)*((-1/6)exp(-6t) + exp(-1t)) = 4.8*exp(-t) - 0.8*exp(-6t);

Bingo!

Now lets' go back to the general equation and assume that X is the current waveform:

x(t) = A1*exp(-1t) + A2*exp(-6t);
i(t) = A1*exp(-1t) + A2*exp(-6t).

Initial conditions: iL(0-) = iL(0+) = 4A. Current derivative:

vL = LdI/dt; voltage across the inductor
dI(0+)/dt = vL(0+)/L = 0.

The system of equations:

4 = A1 + A2;
0 = -A1 - 6*A2;
------------------> A2 = -0.8, A1 = 4.8.

Current waveform equation is

i(t) = 4.8*exp(-t) - 0.8*exp(-6t);

Here we are.

PS! I sincerely hope that this is not a coincidence and this reasoning can be applied to all circuits.

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  • \$\begingroup\$ Once in a very rare moon may I need to write nothing more than a pointer and then follow it up with nothing more than a hint suggesting no particular priority of one direction over another. You are one of those rare moon days! I will remember and I do expect a great deal from you. Nicely done! +1 to both question and answer. \$\endgroup\$ – jonk Aug 23 '19 at 4:18
  • \$\begingroup\$ The now-obvious insight you sought and almost had at the tip of your tongue was already present in your very question. You were so close when you wrote: "which variable should I choose: voltage or current?" I had only time enough then to point in a convenient direction. But when you asked again in a comment, I suspected you would need only a slight nudge. And you grabbed it from there, entirely on your own! You own this, now!! \$\endgroup\$ – jonk Aug 23 '19 at 4:36

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