# Nodal analysis help

#### prodisoft

Joined Oct 17, 2015
7
Hi all, i am trying solve the 5.15 exercise, is for modeling and analisis of dymanic sistems, i have a dude using nodal analisis i have I1, this involves L1, Va, R1?

#### shteii01

Joined Feb 19, 2010
4,644
So...
What have you got so far?

#### prodisoft

Joined Oct 17, 2015
7
I have this but i have stuck on I1

#### shteii01

Joined Feb 19, 2010
4,644
The voltage across inductor is:

so Va-V1=L*dI1/dt
I1 then is:
I1=integral of (Va-V1)/L

#### prodisoft

Joined Oct 17, 2015
7
Thanks a lotfor your help, my dude now is and where i use the Resistor in the analisis?

#### shteii01

Joined Feb 19, 2010
4,644
Thanks a lotfor your help, my dude now is and where i use the Resistor in the analisis?
Which one is your reference node?

#### prodisoft

Joined Oct 17, 2015
7
The black point under C

#### WBahn

Joined Mar 31, 2012
30,038
Your equations are a hash from the get go. You need to learn to track and check your units.

Your very first equation has "R + L" in the denominator. So you are trying to add resistance to inductance. Resistance has units of ohms, which are equivalent to "volts/amp". Inductance has units of "volt·sec/amp" which is equivalent to "Ω·s". They are incompatible and, had you checked units, you would have known that before wasting any more time on anything further.

Similarly, in your next two equations you have current equal to voltage divided by capacitance (which yields V²/A·s, which is no where close to A) and in the last one you have voltage divided by inductance (which yields A/s).

Most of the mistakes you make will mess up the units and, if you bother to check them, you can catch them very early. If you don't check them ....

#### prodisoft

Joined Oct 17, 2015
7
Yes i see it. sorry for that mistake, i used it just for indicate, you have the reason is confuse.

In my node i have 3 currents, I1, I2, I3

My problem is I1, i think R is part of I1, but how I involve R1 to form part of the equation for I1

#### WBahn

Joined Mar 31, 2012
30,038
In your equation for I1, you have (Va-V1), but that is NOT the voltage across the inductor. Va is the voltage at the top-left node relative to the bottom-left node. You need the voltage at the top-left node relative to your chosen common. You are correct in your supposition that you need to get R1 involved somehow. What is the voltage at the bottom-left node relative to your common in terms of R and I1?

#### prodisoft

Joined Oct 17, 2015
7
In your equation for I1, you have (Va-V1), but that is NOT the voltage across the inductor. Va is the voltage at the top-left node relative to the bottom-left node. You need the voltage at the top-left node relative to your chosen common. You are correct in your supposition that you need to get R1 involved somehow. What is the voltage at the bottom-left node relative to your common in terms of R and I1?
for L1 Something like this? (−I1R+Va−V1).

#### WBahn

Joined Mar 31, 2012
30,038

#### prodisoft

Joined Oct 17, 2015
7
Then for I1, I have

I1=integral of (−I1R+Va−V1)/L?

#### WBahn

Joined Mar 31, 2012
30,038
Then for I1, I have

I1=integral of (−I1R+Va−V1)/L?
I think it's L1, isn't it (which is important since you have two inductors in the circuit).