# RLC superposition

Discussion in 'Homework Help' started by Jess_88, Apr 30, 2011.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
hey guys

wondering if my analysis procedure is correct for the cct.

I need to fined io for the following cct using superposition.

so I gather I'm soposed to analyse the current io with resect to each individual source in the circuit, by zeroing remaining sources.

I'm a little unsure of the "zeroing" of each source and if it causes a s/c or o/c.

These are my circuits I determined by zeroing the sources. do they look right?

This circuit contains a dc source. therefor the inductor =s/c cap = o/c

Have I misinterpreted the zeroing sources? or dose this look ok?
thanks a bunch guys.

ps.
I apologise if I am posting so manny questions in this forum. I'm just trying to get as much help over the range of questions I have to do, while people smarter than myself are online (jegues).
If i'm posting to manny equation, let me know

2. ### jegues Well-Known Member

Sep 13, 2010
735
43
Everything looks fine except for the first case where you are zeroing all other sources aside from the 24V DC source.

You still need the 1 ohm branch with the capacitor, as well as the 2 Henry inductor and 4 ohm resistor.

Are you assuming steady state for the capacitor/inductor? Is that why you aren't indicating that branch?

I'm not entirely sure if you can do that. (Is there a reason you think you can?)

Most of what I've learnt about circuits has been from other members on this fourm, and I myself am still learning.

Jess_88 likes this.
3. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
In my lecture notes it assumes steady state when analysing the DC source for a similar problem. The similar question to fine the voltage across a resister, results in three separate voltages which can be added together due three different frequencies (dc w =0), but is expressed at V = V1 + V2 + V3 (eg: V(t) = 1 + 26sin(3t +15) + 7.8cos(5t + 30))

hope that helps... and thang you again for the help

4. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Just checking if my method of finding Io for each of the circuits I drew is ok.

first one is easy(24/6)

for the second I have
It = ((1-j2) + (2//(4+j6)))/10<-120
io = It(2/(2+6)+j6)

third cct
Zt = [(2//(1-j2))]//(4+j6)
Io = [(2//(1-j2)) / Zt] x (6<0)

look ok?