Question

Can some one help me determine whether this circuit is a 2nd order series or parallel or General RLC

Answer 1

and can you help me determine the characteristic equation of i_L (t) for t > 0

Answer 2

When the switch closes the inductor will resist any change so the current through it will not change and will continue to be 16 Volts/8 ohms = 2 amps

The sudden increase in current will immediately be diverted though the capacitor, which will be about 3 amps and probably a little less because some will be forced back into the 5 ohm resistor.

The capacitor will however resist a sudden change in voltage, similar to how the inductor resisted a sudden change in current, so it will be about 16*3/8 = 6 volts immediately after the switch closes.

The voltage across the inductor will increase to about 16-3=13 Volts. We use 3 here because the current across the 3 ohm resistor will go from 2 to -1 amps when the switch closes creating a -3 voltage across this resistor.

Once the current through the capacitor stops changing, there will be no current going through it and so since an ideal inductor has zero resistance, all 3 amps plus the original 2 amps will go through the inductor. This means in stead-state there will be 5 amps going though the inductor.

See the attached simulation of these results. Notice that at 0.5 seconds I flip the switch and then the voltage across the inductor, current across the capacity and current across the 3 ohm resistor all change immediately. All other voltages and currents experience a lag.

Hope this helps.

Answer 3

If you want to analyze this analytically, look at this https://www.youtube.com/watch?v=LmWvJF8yop8

Answer 4

Finally, here are some specifics to solve this problem analytically.

You wanted to know \(i_L(t)\) for \(t>0\). I'll provide the equations to do this but will solve for \(v_c(t)\) instead and will let you solve for the inductor current. The process will be exactly the same.

First lets find two equations for the so-called "state" equations involving the inductor and capacitor.

KCL at the capacitor node gives us:

$$
1)~~~~-Cv'_C(t)-\left({{v_C(t)-V}\over{R_2}}\right )+i-i_L(t)=0
$$

KVL in the loop containing the inductor gives us:
$$
2)~~~~Li'_L(t)-v_c(t)-R_1(i-i_L(t))=0
$$
Where \(i\) is the constant current source,\(V=16~\text{volts}\) ,\(R_1=3~\Omega\) and \(R_2=5~\Omega\).

Using \(1)\),
$$
3)~~i_L(t)=-Cv'_C(t)-\left({{v_C(t)-V}\over{R_2}}\right )+i\\
i'_L(t)=-Cv''_C(t)-{{v'_C(t)}\over {R_2}}
$$
Plugging these into \(2)\) gives us after some algebra:
$$
v''_C(t)+\left ({{L+CR_1R_2}\over{LCR_2}}\right)v'_C(t)+{{R_1+R_2}\over{LCR_2}}v_C(t)={{VR_1}\over{LCR_2}}
$$
After plugging in values, we get
$$
v''_C(t)+55v'_C(t)+1600v_C(t)=9600
$$
The solution is underdamped (because \(\alpha <\omega_n\)):
$$
4)~~ v_C(t)=e^{-27.5t}\left(k_1\cos29t+k_2\sin29t)\right )+k_3
$$
We use initial conditions to find our constants:
$$
v_C(\infty)=k_3=6~\text{volts}\\
v_C(0)=6~\text{volts}\\
i_L(0)=2~\text{Amps}
$$
So using equation \(3)\) we find:
$$
v'_C(0)=600
$$
Then we equate this with \(v'_C(0)\) in equation \(4)\):
$$
600=-27.5k_1+29k_2\
$$
Two constants, we need another equation:
$$
v_C(0)=6=k_1+k_3
$$
This gives us our final solution for the voltage \(v_C(t)\) for \(t\ge0\):
$$
v_C(t)=20.7e^{-27.5t}\sin29t+6
$$
If you graph this function, say in Excel, you'll find that the curve exactly matches the simulated results above.

The process for finding \(i_L(t)\) uses these same equations and follows this exact same process.

The simulator I used is free and can be found at http://qucs.sourceforge.net/.

Hope this helps you.

Answer 5

I went ahead and solved all equations for each current and voltage. I have not however included their derivations. I've plotted the equations here in Excel--the formulas are included in their respective tabs.

Let me know if you have any questions.

I hope this inspires you to arrive at the same solutions I did.