Simple circuit.

  • Engineering
  • Thread starter Would
  • Start date
  • #1
4
0

Homework Statement


practiceprob.jpg



Homework Equations


I am trying to do some self study here before I take my circuits course. I am stuck on this problem for some reason.. The equation I am coming up with for the node is : 6 - Vo/2 - Io/4 - Vo/8 = 0.. After getting at equation I cant seem to come to the answer. Do I need to do another equation here? Thanks


The Attempt at a Solution

 

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
Welcome to PF Would!

Homework Statement


practiceprob.jpg



Homework Equations


I am trying to do some self study here before I take my circuits course. I am stuck on this problem for some reason.. The equation I am coming up with for the node is : 6 - Vo/2 - Io/4 - Vo/8 = 0.. After getting at equation I cant seem to come to the answer. Do I need to do another equation here? Thanks


The Attempt at a Solution


Ohm's law for the 2 ohm resistor tells you that i0 = v0/(2 Ω). That's your second equation. Now you have everything in terms of only one unknown, and you can therefore solve the system.
 
  • #3
4
0
Thanks a lot for the help! I must be doing some simple math wrong here.. but im getting : 6 - Vo/2 - Vo/4 - Vo/8 = 0 then multiplying all of that by 8 to get 48- 4Vo - 2Vo - Vo = 0 then ultimately getting Vo = 6.85V but that is wrong... What am I doing wrong there? Should I be assuming that the current in across the 8 ohm resistor is moving upward and therefore making the equation 48 - 4Vo - 2Vo + Vo = 0? It works out that way but if it is like that could you tell me why? Since it is not marked i assumed it was a voltage drop across the 8 ohm resistor..
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
Thanks a lot for the help! I must be doing some simple math wrong here.. but im getting : 6 - Vo/2 - Vo/4 - Vo/8 = 0 then multiplying all of that by 8 to get 48- 4Vo - 2Vo - Vo = 0 then ultimately getting Vo = 6.85V but that is wrong... What am I doing wrong there? Should I be assuming that the current in across the 8 ohm resistor is moving upward and therefore making the equation 48 - 4Vo - 2Vo + Vo = 0? It works out that way but if it is like that could you tell me why? Since it is not marked i assumed it was a voltage drop across the 8 ohm resistor..

The third term in your equation is simply wrong. In your original KCL equation, this term was i0/4, which was correct. Then we determined that i0 = v0/2. Hence, i0/4 = (v0/2)/4 = v0/8. The third term (in red) should therefore be v0/8.
 
  • #5
4
0
Wow I feel stupid after looking back at that. Thank you for the help.
 

Related Threads on Simple circuit.

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
4K
Replies
2
Views
1K
Replies
5
Views
3K
  • Last Post
Replies
6
Views
3K
W
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
1
Views
4K
Replies
3
Views
8K
  • Last Post
Replies
1
Views
5K
Replies
8
Views
3K
Top