How to deal with dc source on ac circuits?

Click For Summary
SUMMARY

The discussion focuses on analyzing AC circuits with a DC source using the superposition theorem and phasor analysis. The voltage across a 3-ohm resistor was calculated to be 3.2cos(6t+126.87) using mesh analysis. Participants clarified that the DC current source can be ignored in AC analysis, treating it as an open circuit, and emphasized the importance of using phasor impedances to formulate KCL equations for AC circuits. The final approach involves simplifying the circuit to three parallel elements for further analysis.

PREREQUISITES
  • Understanding of superposition theorem in circuit analysis
  • Proficiency in mesh analysis for circuit calculations
  • Familiarity with phasor representation in AC circuits
  • Knowledge of KCL (Kirchhoff's Current Law) application in circuit analysis
NEXT STEPS
  • Study phasor analysis techniques for AC circuit analysis
  • Learn about the superposition theorem in the context of AC circuits
  • Explore the behavior of inductors and capacitors in steady-state AC conditions
  • Investigate the implications of ignoring DC sources in AC circuit analysis
USEFUL FOR

Electrical engineers, circuit designers, and students studying AC circuit analysis who need to understand the interaction between DC sources and AC circuits.

Secre
Messages
4
Reaction score
2
Homework Statement
Find the steady state voltage v.
Relevant Equations
v(t) = V(dc) + V(ac)
IMG_20210526_221720.jpg

We are tasked to solve for the steady state voltage v using superposition theorem. I start with solving with voltage source using mesh analysis and got the current flowing through the 3 ohms resistor and get the first value of the voltage which is 3.2cos(6t+126.87). Now my problem is I am confused on how to deal with the other source since we're in ac analysis. I tried solving it with dc analysis but I think it's wrong.
 
Physics news on Phys.org
Have you started using Phasors yet or analyzing AC circuits with a single fixed frequency?
 
Yes
 
Great! The 4A current source looks like a/an __________ for AC, so you can probably use the Phasor impedances to write the AC KCL equation(s)...
 
Should I assume w= 0 rad/s for the current source since I'll be needing it to convert it into phasor circuit?
 
Well, I'd just ignore the DC current source for the AC analysis, since an ideal current source has infinite output resistance (looks like an open circuit). Can you write the Phasor KCL equation for the top middle node (since the DC current source is not an open)?
 
IMG_20210526_232911.jpg
 
To get the contribution to v from the 4 amp source just start by short circuiting the voltage source. Now in the steady state the inductors with be short circuits and the capacitor will be an open circuit. You are then left with three parallel circuit elements (one current source and two 4 ohm resistors.) I think you can take it from there.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
5K
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
6K