| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Jun 2007 21:52:12 IST
|
|
|
I am getting confused about how to measure the potential in a closed circuit which is a complex one i.e., how to apply the direction in which the current was flowing. i find it easier for only the simple circuits. there is a question i came through in which there is a cube and the potential is to be found out at two given points but in that question i am failing to apply kvl. please suggest me a simple way to proceed
|
|
|
|
|
|
|
|
i do not think there is any other way except KVL....lets see what the experts say
|
rate me if you find this useful |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Jun 2007 21:37:53 IST
|
|
|
See understand this whenever you are moving along the direction of current, you are actually moving along a drop in potential. If in moving along a closed path you find a resistance, check whether you are moving along the current. If its so, you have a term -iR in your KVL equation. If you are moving opposite to the current, you have iR. For a capacitor if you encounter the +ve plate first write -q/C. If you encounter the negative plate first write q/C.
|
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 Jun 2007 22:35:38 IST
|
|
|
applly the concept of equipotential junction inthe case of cube .and also use symmetry property.
|
B.Tech CSE, ISMU |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jun 2007 00:14:43 IST
|
|
|
this thing u can only understand by examples.. try ex. in h.c. verma i think its given ther
|
Sure i am that this day we are the masters of our fate, that the task set before us is not above our strengths; that its pangs and toils are not beyond our endurance. As long as we have faith in our own cause and an unconquerable will to win, victory will not be denied us.
- Winston Churchill |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jun 2007 01:17:45 IST
|
|
|
use symmetry and distribute current like that and then solve it?   
|
yugank
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jun 2007 16:37:02 IST
|
|
|
Pick up H C Verma and read each and every line carefully. Solve its solved problems and analyze the approach.
1st of all give currents to the different meshes and then make equations at junctions. When moving in a circuit a resistor is encountered and if the direction of current is along the motion then there is a potential drop, so add -iR to the equation and vice-versa.
|
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Jun 2007 23:51:29 IST
|
|
|
hey, for complex circuits use nodal analysis .....................
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|