Very simple superposition circuit

  • Context: Engineering 
  • Thread starter Thread starter James889
  • Start date Start date
  • Tags Tags
    Circuit Superposition
Click For Summary
SUMMARY

The forum discussion centers on calculating the current in a superposition circuit. The user correctly applies the superposition theorem by first removing one voltage source and calculating the current through the circuit with the remaining source. The total resistance is determined to be 150 ohms, leading to a calculated current of approximately 0.067A from the first source and 0.033A from the second source. The final current through the resistor is confirmed to be 50mA, validating the user's calculations.

PREREQUISITES
  • Understanding of superposition theorem in electrical circuits
  • Basic knowledge of Ohm's Law
  • Familiarity with series and parallel resistor combinations
  • Ability to analyze simple electrical circuits
NEXT STEPS
  • Study the application of the superposition theorem in more complex circuits
  • Learn about Kirchhoff's laws for circuit analysis
  • Explore the use of simulation tools like LTspice for circuit verification
  • Investigate the impact of varying resistor values on circuit behavior
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and professionals needing to analyze and troubleshoot superposition circuits.

James889
Messages
190
Reaction score
1
Hi,

I have didn't have an answer to this, so i need someone to tell me whether my work is correct or not.

This is the circuit:
[PLAIN]http://img251.imageshack.us/img251/4081/circuit1.png

I need to determine the current I.

Remove the first source and calculate the current provided by the other.
[PLAIN]http://img255.imageshack.us/img255/1665/circuit11.png


The total resistance is 150ohm [tex]100 + \frac{100 \cdot 100}{100+100}[/tex]

I = 10/150 ~0.067A

And the current through the resistor is:

[tex]\frac{100}{200} \cdot 0.067 = 33,5mA[/tex]


Do the same with the other source
[PLAIN]http://img69.imageshack.us/img69/9727/circuit12.png

I = 5/150 ~0.033A
[tex]\frac{100}{200} \cdot 0.033 = 16,5mA[/tex]


So the current I should be 50mA
 
Last edited by a moderator:
Physics news on Phys.org
yes correct
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K