# Node Analysis with frequency domain circuit...Help

Discussion in 'Homework Help' started by bserrato86, Apr 19, 2009.

1. ### bserrato86 Thread Starter New Member

Apr 18, 2009
2
0
I need some help understanding where I messed up on an assessment problem I was doing. I took a pic with my phone but it didn't come out so well.

Is=10cosωτ
Vs=100sinωτ

R(next to current source on the left)=5Ω
C=9μF
L=100μH
R(next to voltage source on the right)=20Ω

ω=50κ
--------------------------------------------------------
1st) I changed everything to it's proper form...

Is=10 angle(0°)
Vs=100 angle(-90°)
R=5Ω
C=-j2.2
L=j5
R=20Ω
---------------------------------------------------------
2nd) I got the equation for node analysis...

V-[100 angle(-90)] + V + V + V - 10 = 0
20 j5 -j2.2 5

----------------------------------------------------------
3rd) Did some math...

V + V + V + V -10-[5 angle(-90)]= 0
20 j5 -j2.2 5

.05s + (-j.2)s + (j.45)s => .05s + (j.25)s => [.26 angle(78.7)]s =>

3.84 angle(-78.7) = Z

so............

I=V => 15 angle(-90) = ______v______
Z 3.84 angle(-78.7)

V= 57.6 angle(168.7)
------------------------------------------------------------
Problem...

31.62 angle(−71.57◦)
v = 31.62 cos(50,000t − 71.57◦)V

Where did I go wrong? Also it has to be nodal analysis.

File size:
77.4 KB
Views:
27
2. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
Your attachment is extremely fuzzy. Can you provide a clearer image?

hgmjr

3. ### PRS Well-Known Member

Aug 24, 2008
989
36
I hope this helps. Using < for the angle, I would write the nodal equation like this:

(V-100<0)/20 + V/j5 + V/-j2.22 +V/5 = 10<0

This is the current through the 20ohm resistor plus the current through the inductor plus the current through the capacitor plus the current through the 5 ohm resistor equals the current from the current source.

Good luck! 