Engineering Solving Current in a Circuit: How Many Equations?

  • Thread starter Thread starter magnifik
  • Start date Start date
  • Tags Tags
    Circuit Current
AI Thread Summary
When analyzing a circuit, the placement of current inside or outside the loop affects the number of equations needed. Using current outside the loop requires four loop equations, while placing it inside reduces the requirement to three. The current through a specific resistor remains the same regardless of the current's position in the loop. Combining impedances can simplify the analysis, allowing for fewer loops when calculating currents. Understanding whether the voltage source is DC or AC is crucial for determining the type of analysis to perform.
magnifik
Messages
350
Reaction score
0
I have a circuit I need to analyze shown in the figure immediately following:
24cz9dd.png


What I want to know is, if I put the current inside the loop rather than outside (as in the picture below), would I get the same answer when trying to solve for the second current (indicated by the arrow on the rightmost resistor)?
35ddmon.png


When analyzing with the current on the outside of the loop, I need 4 loop equations, correct?...
because I have the outer loop, the right loop, the left loop, and the loop on the inside of the box with the inductor and resistor.

But when analyzing with the current inside the loop, I only have 3 equations.

I am not asking for help with solving, just in understanding.
 
Physics news on Phys.org
magnifik said:
I have a circuit I need to analyze shown in the figure immediately following:
24cz9dd.png


What I want to know is, if I put the current inside the loop rather than outside (as in the picture below), would I get the same answer when trying to solve for the second current (indicated by the arrow on the rightmost resistor)?
35ddmon.png


When analyzing with the current on the outside of the loop, I need 4 loop equations, correct?...
because I have the outer loop, the right loop, the left loop, and the loop on the inside of the box with the inductor and resistor.

But when analyzing with the current inside the loop, I only have 3 equations.

I am not asking for help with solving, just in understanding.

The current I1 in both images is exactly the same, it is simply passing through that one resistor. (So yes, the 2nd current you've indicated on the drawing should be the same)

You should only have 3 loop equations either way.
 
Following up on what jegues stated, if you're just looking for the current in the rightmost resistor, you can combine the impedances of the resistor/coil combination into a single (complex) value and you're left with two loops instead of three.

I suppose I should have asked, is the voltage source a DC source or an AC source? (Are you performing a transient analysis or a steady-state AC analysis)?
 
gneill said:
Following up on what jegues stated, if you're just looking for the current in the rightmost resistor, you can combine the impedances of the resistor/coil combination into a single (complex) value and you're left with two loops instead of three.

I suppose I should have asked, is the voltage source a DC source or an AC source? (Are you performing a transient analysis or a steady-state AC analysis)?


steady state analysis
 
I1 would be the current going through the entire loop of the left-hand side of the circuit, right? And I2 is the current going through the right side of the circuit
 
See the attached figure. I've indicated two loop currents. Note that I've assumed that the parallel RL combo in the first loop is taken as a single component.

It looks like you'll have to write a KVL equation for each loop. The controlled voltage source (k*I1) in loop 2 makes it interesting.
 

Attachments

  • Fig4.jpg
    Fig4.jpg
    9.2 KB · Views: 367
Last edited:
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...

Similar threads

Replies
1
Views
881
Replies
4
Views
2K
Replies
7
Views
3K
Replies
5
Views
2K
Replies
5
Views
2K
Replies
3
Views
2K
Replies
5
Views
2K
Back
Top