Circuit with capacitors and inductors

AI Thread Summary
In a long-connected circuit with capacitors and inductors, the inductor's voltage is zero, allowing for nodal analysis by eliminating inductors. Capacitors in steady state have constant voltage and do not pass current, leading to their removal and the shorting of inductors, leaving only resistors. The discussion clarifies that current sources cannot be summed if they are not in parallel, emphasizing that the 10A source is the sole supply for the parallel resistors. The 2A source does not affect the circuit because the ideal 10A source dictates the current flow, ensuring all 10A is utilized. The current flowing through the components is confirmed to be 10A to the left, illustrating the behavior of ideal current sources.
eurekameh
Messages
209
Reaction score
0
2r2cbh5.png

The question states that the circuit has been connected for a very long time. I'm assuming that then, the current will remain constant and thus the voltage for the inductor is vL = Ldi/dt = 0. Thus, using nodal analysis, I was able to solve (a) by eliminating all of the inductors from the circuit. For (b), however, there is a capacitor for which I do not know how to handle, although I would guess that a mesh analysis would be involved from the looks of the current sources.
 
Physics news on Phys.org
When capacitors reach steady state in a DC circuit they reach a constant voltage and pass no current. You can remove capacitors and short out inductors...
 
I have removed the capacitors and shorted out the inductors. All that remains are the resistors. Would it be valid to sum the current sources and put the 12 A source to the left of everything and use current division?
 
eurekameh said:
I have removed the capacitors and shorted out the inductors. All that remains are the resistors. Would it be valid to sum the current sources and put the 12 A source to the left of everything and use current division?

No, you can't sum the current sources as they are not in parallel. But you don't need to: the 10A supply is the only supply to the 2k and 5k resistors which are in parallel (now that the inductor is shorted). Apply current division.
 
Why doesn't the 2 A source have no added effect? I would assume from intuition that it would contribute to the current i,x going through the 5 k,ohm resistor.
 
Last edited:
eurekameh said:
Why doesn't the 2 A source have no added effect? I would assume from intuition that it would contribute to the current i,x going through the 5 k,ohm resistor.

The 2A source can't have any effect on those components because the 10A current source will pass 10 amps, and ONLY 10 amps no matter what. An ideal source cannot do anything but exactly what it's specified to do. If an ideal source is putting 10A into a node, you better believe that all 10A are going somewhere! :smile:

If you were to draw a current arrow on the bottom rail that is below the 10A source, what current value would you label it with? Where does that current come from ?
 
10 A going to the left?
 
eurekameh said:
10 A going to the left?

Yes. Where does that current come from (through which components must it flow)?
 

Similar threads

LC oscillations - two capacitors and an inductor
Replies
15
Views
522
Current through an inductor as a function of time
Replies
10
Views
310
Determining Inductance L in an LC Circuit
Replies
5
Views
2K
Sign conventions for Kirchhoff's loop rule
Replies
7
Views
1K
Symmetrical parallel inductors, different currents? State Preparation
Replies
7
Views
1K
Self-inductance of LC circuit given rate of capacitor discharge.
Replies
6
Views
861
Current in RLC Series Circuit: Need Help
Replies
2
Views
1K
Calculation of current in driven series RLC circuit
Replies
8
Views
2K
Combined tensions of an RLC series circuit with AC current
Replies
3
Views
830
Potential Drop Across 15-mH Inductor at t=0?
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top