What are the equations for calculating loop currents in a circuit?

  • Thread starter Thread starter theuniverse
  • Start date Start date
  • Tags Tags
    Circuit Current
Click For Summary
SUMMARY

The discussion focuses on calculating loop currents in a circuit using Kirchhoff's laws. The initial equations presented by the user were incorrect, leading to erroneous current values of I_1=1.43A, I_2=-0.71A, and I_3=-1.07A. The correct equations, which yield the accurate currents of I_1=3.55A, I_2=-1.98A, and I_3=-2.98A, are provided as follows: 1) 10V-5I_1-2I_1-5I_3=0, 2) 10V-5I_1-5I_3-3I_3-20V-4I_3+4I_2=0, and 3) 4I_2-4I_3+2I_2=0. The key takeaway is the importance of correctly accounting for all currents in each resistive component of the loop.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Ohm's Law (V=IR)
  • Basic knowledge of circuit analysis techniques
  • Ability to solve simultaneous equations
NEXT STEPS
  • Study Kirchhoff's Laws in depth
  • Practice solving circuit problems using mesh analysis
  • Learn about the impact of resistive components on circuit behavior
  • Explore advanced circuit simulation tools like LTspice
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in circuit analysis and troubleshooting.

theuniverse
Messages
53
Reaction score
0

Homework Statement


Find the Loop currents that are shown in the network of the figure below.

Homework Equations


V=IR

The Attempt at a Solution


I know that I need to create 3 equations since there are three unknown currents and then solve for them so here is what I got.
1) 10V-5I_1-2I_1=0
2) 10V-5I_3-3I_3-20V-4I_3+4I_2=0
3) 4I_2-4I_3+2I_2=0

When I solve for the currents I get
I_1=1.43A
I_2=-0.71A
I_3=-1.07A

While the correct answers should be 3.55A, -1.98A, -2.98A. I suspect that my equations are incorrect but I can't come up with any equations that give me the right answer or make any sense.

Thanks,
 

Attachments

  • Circuit.jpg
    Circuit.jpg
    15 KB · Views: 444
Physics news on Phys.org
As you 'traverse' each resistive component in a given loop, you need to take into account all the currents flowing through that component. So, for example, the voltage drop across your 5Ω resistor is (I1 + I3)*5Ω.
 
Yea that was the concept I couldn't fully understand... I get it now.
In case anyone else is interested here are the equations that actually work:
1) 10V-5I_1-2I_1-5I_3=0
2) 10V -5I_1-5I_3-3I_3-20V-4I_3+4I_2=0
3) 4I_2-4I_3+2I_2=0

gneill You've been a great help as usual! Thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
1K