Solve Nodal Analysis Differential Equation for I1 with Reference Point under C

[Jair Gutierrez]

Homework Statement

Using nodal analisis we need obtain the differential equation

2. The attempt at a solution

For I1 and take the point under C as reference, but i am stuck on it

 

[NascentOxygen]

Hi Jair Gutierrez.

The next step here is to add those 3 currents, and equate the result to 0; this gives you your second order DE.

You don't have the right expression for I₁, the current in an inductor.

 

[gneill]

Is Va entirely arbitrary or does it have some particular form? I'm assuming it's a source rather than the desired output from some initial value conditions of the circuit... the problem statement is rather vague.

 

[Jair Gutierrez]

Hi thanks for your anwers, yes I1 is an inductor,
I1=integral of (Va-V1)/L, but in all this equations where i use the resistor R, i think R is part of I1, but i don't know how write the equation in this case.

Here is the original exercise, 5.15

 

[Jair Gutierrez]

In my node i have 3 currents, I1, I2, I3, i think R is part of I1, but how I involve R1 to form part of the equation for I1

 

[gneill]

The potential at the left side of L1 is not Va since, as you say the resistor R is between it and the reference node. There will be a potential drop across R due to $I_1$. So the potential across L1 is going to be something like: $(-I_1 R + V_a - V_1)$.