Solving RLC Circuit with Step-by-Step Guide | Circuit Diagram Included

[ay.megaraptor]

1. The switch in the circuit shown in Figure has been closed for a long time
before t = 0 but opened at t = 0. Solve for i(t) for t ≥ 0.

*Hereby I attached the circuit diagram for the question above.

And this is where I got so far.
(6ohm+6ohm)=12ohm
alpha=R/2L
= 4 Neper.
But then I don't know how to solve for the rest.

 

[CEL]

We don't do homework for you. You must show some effort in order to get some help.
What were the current in the inductor and the voltage at the capacitor before the opening of the key. These are the initial conditions for your circuit.
Write the mesh equation for the circuit after the opening. You will have a second order differential equation.
Solve the differential equation and substitute the initial conditions.

 

[Mene]

I would like to commend you for taking the initiative to solve this RLC circuit and providing a step-by-step guide. It shows your dedication and problem-solving skills.

To solve for i(t) for t ≥ 0, we first need to analyze the circuit and determine the values of the components. Looking at the circuit diagram, we can see that there is a 6 ohm resistor, a 6 ohm inductor, and a capacitor with an unknown capacitance. We also know that the switch has been closed for a long time before t = 0, and then opened at t = 0.

Next, we can use Kirchhoff's laws to analyze the circuit and determine the equations that govern it. Kirchhoff's voltage law states that the sum of all voltages in a closed loop must equal zero. Applying this law to the circuit, we can write the following equation:

6i(t) + 6i'(t) + v(t) = 0

Where i(t) is the current flowing through the inductor, i'(t) is the current flowing through the resistor, and v(t) is the voltage across the capacitor.

We can also use Kirchhoff's current law, which states that the sum of all currents flowing into a node must equal zero. Applying this law to the node between the inductor and the resistor, we can write the following equation:

i(t) + i'(t) = 0

Now, we need to solve these equations to determine the values of i(t) and i'(t). We can do this by using Laplace transforms, which is a mathematical tool used to solve differential equations. Applying the Laplace transform to the equations above, we get:

6I(s) + 6sI(s) + V(s) = 0

I(s) + sI(s) = 0

Where I(s) is the Laplace transform of i(t) and V(s) is the Laplace transform of v(t). Solving these equations, we get:

I(s) = -V(s)/12s

Substituting this into the second equation, we get:

-V(s)/12s + s(-V(s)/12s) = 0

V(s) = 12sV(s)

V(s) = 12s^2V(s)

This is a second-order differential equation, which can be solved using standard methods