RLC Parallel Circuit: Analyzing Current & Voltage at t=0-

Click For Summary
SUMMARY

The forum discussion focuses on analyzing an RLC parallel circuit at the moment just before and after a current source is introduced at t=0. At t=0-, the inductor current i(L) is 0A, and the voltage across the capacitor V(c) is -20V due to the steady-state condition established by the voltage source. The participants clarify that at t=0+, both the inductor current and capacitor voltage remain unchanged, leading to a voltage across the resistor VR of 0V, as it is the sum of the capacitor voltage and the voltage source. The discussion emphasizes the importance of applying Kirchhoff's Voltage Law (KVL) and nodal analysis to derive the circuit's response.

PREREQUISITES
  • Understanding of RLC circuit theory
  • Knowledge of Kirchhoff's Voltage Law (KVL)
  • Familiarity with nodal analysis techniques
  • Basic concepts of steady-state and transient analysis in circuits
NEXT STEPS
  • Study the application of Kirchhoff's Voltage Law (KVL) in RLC circuits
  • Learn about nodal analysis for complex circuit configurations
  • Explore transient response analysis in RLC circuits
  • Investigate the effects of initial conditions on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing RLC circuits and their transient responses will benefit from this discussion.

jisbon
Messages
475
Reaction score
30
Homework Statement
Find the desired currents and voltages for the circuit below.
Relevant Equations
-
1598619583802.png

1598619595419.png

These are my attempts at doing this question, and I was wondering if I am correct so far.
At t= 0-, i(L) will be 0A, since the capacitor acts as a open circuit.
However, I'm not sure why V(c) at t=0- will be -20V as given by the answers. Won't it be 0V?

Moving on, since current in inductor and voltage in capacitor does not change immediately, the values for t=0+ will be the same as t=0-. (Again, I'm not sure why it is -20V for the capacitor, could someone explain to me?)

I'm not sure how to proceed, but if it was me, I will probably transform the circuit? to:

1598619861097.png

Taking KVL on the left loop,
-80+20i1+ Vc+20=0 --(1)
Taking KVL on right loop,
-20-Vc+40iL+1/L (integerate voltage) -- (2)

I also know Vc = 1/C (integerate i1-iL)

Now the thing is am I making this more complicated than I should? or is this correct so far?
Thanks
 
Physics news on Phys.org
jisbon said:
I'm not sure why V(c) at t=0- will be -20V
The voltage source was connected in the circuit for a sufficiently long time. So the circuit was in steady state before the current source was switched at t=0.

You are right about the inductor current and capacitor voltage at t=0+.
Based on this conclusion, what can you say about VR at t=0+?
 
cnh1995 said:
The voltage source was connected in the circuit for a sufficiently long time. So the circuit was in steady state before the current source was switched at t=0.

You are right about the inductor current and capacitor voltage at t=0+.
Based on this conclusion, what can you say about VR at t=0+?
1598623772757.png

Regarding the voltage, what I do know (from theory is that)it will be a open circuit, but I'm still unclear why it is -20V in this case.As for the voltage ofthe resistor, won't i need to calculate what is the current running through it using the analysis I listed above? Or am I missing something?
 
jisbon said:
Regarding the voltage, what I do know (from theory is that)it will be a open circuit, but I'm still unclear why it is -20V in this case.
What is the potential difference across the capacitor at t=0-? You know there is no current through the capacitor. There is no current source in the circuit yet.
Apply KVL and compare it with the reference polarity.
Also, why have you replaced the inductor with a short circuit in your previous post?
 
jisbon said:
but if it was me, I will probably transform the circuit? to:
That would work if you want to find all of the asked quantities except VR and dVR/dt.
For finding VR and dVR/dt, you need to leave R the way it is, since it behaves differently when clubbed together with the transformed voltage source.
 
  • Like
Likes etotheipi
cnh1995 said:
Also, why have you replaced the inductor with a short circuit in your previous post?
whoops, that was a mistake.

cnh1995 said:
That would work if you want to find all of the asked quantities except VR and dVR/dt.
For finding VR and dVR/dt, you need to leave R the way it is, since it behaves differently when clubbed together with the transformed voltage source.
so if I'm proceeding with the normal circuit, I would need to apply the usual nodal analysis, right?
 
jisbon said:
so if I'm proceeding with the normal circuit, I would need
Nodal analysis and differential equations would help to find the full response of the circuit. But you are particularly asked about the initial and final values of the response. So you can focus only on initial and final values separately in your analysis.

Initial values first:
You are right about the inductor current and capacitor voltage at t=0+.
When the current source is introduced in the circuit at t=0, the inductor current and capacitor voltage don't change at t=0+. But there is an additional current source in the circuit at t=0+.
Where will this current go at t=0+ ?
 
cnh1995 said:
Nodal analysis and differential equations would help to find the full response of the circuit. But you are particularly asked about the initial and final values of the response. So you can focus only on initial and final values separately in your analysis.

Initial values first:
You are right about the inductor current and capacitor voltage at t=0+.
When the current source is introduced in the circuit at t=0, the inductor current and capacitor voltage don't change at t=0+. But there is an additional current source in the circuit at t=0+.
Where will this current go at t=0+ ?
can I assume it goes to the resistor only?
Hence voltage of resistor at t=0+ = 80V?
However, the answers given to me stated that it is 0V :/
 
jisbon said:
Hence voltage of resistor at t=0+ = 80V?
No.
Hint: Voltage across the resistor is the sum of capacitor voltage and 20V source voltage.
You already know the capacitor voltage at t=0- and t=0+.
What does that tell you about the voltage across the resistor at t=0+?

Keep in mind the initial conditions of L and C that you already stated correctly. You need to use them to verify if your assumptions are correct.
 
  • #10
cnh1995 said:
No.
Hint: Voltage across the resistor is the sum of capacitor voltage and 20V source voltage.
You already know the capacitor voltage at t=0- and t=0+.
What does that tell you about the voltage across the resistor at t=0+?

Keep in mind the initial conditions of L and C that you already stated correctly. You need to use them to verify if your assumptions are correct.
Ah okay, since voltage across the resistor is the sum of capacitor voltage and 20V source voltage, and voltage at t=0+ of capacitor =-20V, -20+20=0V

As for di/dt at t=0+, I understand that v=L(di/dt). However in this case, will the voltage across the inductor be 0? (even though there is a 40 ohm resistor across the voltages of capacitor and 20V) and the inductor. I would assume so since the answer for di/dt across inductor turns out to be 0.

Also, thinking about dv/dt across capacitor and resistor,the answer turns out to be 4.
For capacitor, i=C(dv/dt). The question is, why will the current going through the capacitor be 4 in this case? Won't some of the current be distributed to the 20 ohm resistor?
 
  • #11
jisbon said:
Won't some of the current be distributed to the 20 ohm resistor?
Not at t=0+.
jisbon said:
since voltage across the resistor is the sum of capacitor voltage and 20V source voltage, and voltage at t=0+ of capacitor =-20V, -20+20=0V
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
862
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
929
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K