# Initial Conditions and Parallel Resonant Circuit Problem

simulate this circuit – Schematic created using CircuitLab

I already did my own solution but I just want to check if I got everything right, especially in determining the initial conditions. Here is my own understanding:

So, for t<0, the current source and capacitor is an open circuit and the inductor will be a short circuit. Hence, all the current will flow to the short circuited inductor. $$t<0:$$ $$i(0) = 3A;$$ $$v(0) = 30;$$ $$\frac{di (0)}{dt} = \frac{30}{4} = 7.5$$

Now, at t > 0: We could do source transformation with the 30V source and 10 ohm resistor, hence, everything will be in parallel. I combined the 6 A and the 3 A to get 9 A current source and combined the two resistances to get:

$$t>0:$$ $$i(0^+) = 9A;$$ $$R = 10||40 = 8;$$ $$resonant freq. = \frac{1}{√LC} = 5 = 7.5$$ $$α = \frac{1}{2RC} = 6.25$$ $$s_1 = -2.5, s_2 = -10$$ Here we can see that we will have an overdamped response hence our solution would be of the form: $$i(t) = I_f + A_1e^{-2.5t} + A_2e^{-10t}$$

To get A1: $$i(0) = I_f + A_1 + A2$$ $$3 = 9 + A_1 + A_2$$ $$-6 - A_2 = A_1$$

To get A2: $$\frac{di (0)}{dt} = \frac{30}{4} = 7.5 = -2.5A_1 + -10A_2$$ $$7.5 = -2.5(-6-A_2) - 10A_2$$ $$A_2 = 1$$ $$A_1 = -7$$

So my final equation would be: $$i(t) = 9 -7e^{-2.5t} + e^{-10t}$$

Did I do everything right? I feel like my initial conditions analysis is wrong but when I checked using LTspice, im getting the current inductor to be almost 3A for t<0 and 9A for t>0. But when I check for voltage on the node of Vc, im getting 3mV which I don't understand.

## 1 Answer

I'll present here the theoretical solution. For $$\- \infty < t < 0\$$ only the voltage source is present in this RLC parallel circuit, which establishes an initial inductor current of $$\i(0-)=i(0+)= 30 \space V / 10 \space \Omega = 3 \space A\$$. Note also that $$\v(0-)=v(0+)= 0 \space V\$$, since the inductor can be seen as a short circuit for $$\t < 0\$$. So, the circuit can be converted as shown on figure below:

Applying KCL on upper node:

$$-I + \frac{v(t)}{R} + C\frac{d}{dt}v(t) + i(t) = 0$$

Replacing $$\v(t) = L\frac{d}{dt}i(t)\$$

$$LC\frac{d^2}{dt^2}i(t) + \frac{L}{R}\frac{d}{dt}i(t) + i(t) = I$$

The characteristic polynomial is:

$$LCs^2 + \frac{L}{R}s + 1 = 0$$

With roots

$$s_{1,2} = -\frac{1}{2RC} \space \pm \sqrt{\left ( \frac{1}{2RC}\right )^2 - \frac{1}{LC}}$$

As $$\\left ( \frac{1}{2RC}\right )^2 > \frac{1}{LC}\$$, the system is overdamped, with two real and distinct roots $$\s_1 = -2.5\$$ and $$\s_2 = -10\$$ for the current case.

The complete response has the form

$$i(t) = i_f + A_1e^{-s_1t} + A_2e^{-s_2t}$$

where $$\i_f\$$ is the forced response (in this case, $$\9 \space A\$$). The constants $$\A_1\$$ and $$\A_2\$$ can be determined from the initial conditions.

$$\left\{\begin{matrix} i(0) = I + A1 + A2 & (1)\\ \frac{d}{dt}i(0) = -s_1A_1 -s_2A_2 & (2)\\ \end{matrix}\right.$$

Note that $$\v(0) = L\frac{d}{dt}i(0)\$$. As $$\v(0) = 0 \space V\$$ then, $$\\frac{d}{dt}i(0) = 0 \space A/s\$$ Therefore

$$\left\{\begin{matrix} 3 = 9 + A1 + A2 & (1)\\ 0 = -2.5A_1 -10A_2 & (2)\\ \end{matrix}\right.$$

Resolving, $$\A_1 = -8\$$ e $$\A_2 = 2\$$

Finally, the inductor current in Ampere is

$$i(t) = 9 -8e^{-2.5t} + 2e^{-10t}$$

With graph

• I have since edited my sol. since I posted this question a few days ago and I have the same answer as you! I got confused by my i(0) that's why my A1 and A2 was wrong on my first solution.
– user263783
Oct 19 '20 at 12:48