E-StoreHome » Ask & Discuss » Physics. » Electricity « Back to Discussion
Electricity
First read this article before seeing this thread-
http://www.goiit.com/posts/list/0/community-shelf-circuit-solving-was-never-so-easy-67415.htm#332300
I have posted there an example of rearranging circuits, here I am posting an example of solving circuits. U can also post ur doubts in circuits here only and I wud try my best to solve them.
The question is very common i.e., to find current in each branch.
In the second figure, I have applied 0 and V potential to A and B terminals respectively. As a,b,c are connected to A by a wire, A,a,b,c wud be at same potential.
Now the potentials of d,e,f are calculated as we are going from higher to lower potential of the battery, so we wud subtract battery's voltage from V.
Now comes the real application of Nodal analysis. U can consider either 0 or V-10/V-8/V-12 to be at higher potential according to ur own will. It will not alter the answer. Here I am considering 0 to be at lower potential and as current flows from higher to lower potential, the direction of current wud be as shown.
Now apply kirchoff's junction law at B. Means net current entering B = 0.
Now, I = V/R
so,
(V-12-0)/4 + (V-10-0)/4 + (V-8-0)/5 = 0
from here on solving, u wud get, V = 71/7.
Now u know potential difference across every resistor and u can now find current in each branch. 
Comments (23)
@ganesha1991,
Ur question is a simple one. First try to simplify the circuit to maximum possible extent in its given design.
For example, the 2 capacitors shown in first drawing of Figure-1 are in series so in second drawing, I replaced them with a single capacitor of C/2.
Now both the ends of this C/2 capacitor are connected by the same wire, hence the potential across both its ends wud be same(as potential remains constant in a conductor). So it means that potential difference across this capacitor wud be zero and hence it is short circuited or u can say, it is no more there in the circuit.
So, after removing it, U wud get a more simplified circuit shown in third drawing of Figure-1.
Now coming to figure-2, as I have done in my article, I take A to be at 0 potential, B to be at 100 potential and some arbritrary V potential.
And as I have shown in my article, the points connected by same wires wud have same potentials(Drawing-1 of Figure-2). And as U can see, capacitors 1,2,3 have same potential difference across them, so it means they are in parellel as potential remains same in parellel.
Now naming equivalent points with same numbers(drawing-2 of Figure-2) I have redrawn this circuit(drawing-3 of Figure-2).
Now its very easy to solve. And the answer is 3C/4.
Hope U understand it.
@reddevil_2009,
Ur question is pretty simple, so no need to complicate it using nodal or equipotential method.
Just luk at the figure-
2 capacitors are connected to A, 2 are connected to B and one capacitor is connected between those 4. Hope u understand how I have redrawn it.
Now if all the capacitors are of equal capacity, its a Wheatstone Bridge(Hope u know abt wheatstone bridge). Which means the centre capacitor wud be short circuited and the figure is more simplified to the second drawing of the figure.
Now its really easy to solve, and the answer wud be C.
Hope u get it.
..Cant get its answer.
@reddevil_2009,
Both the middle branches wont get short circuited here. See, one of the branch is directly connected to the terminals. So that branch wont get short circuited. The other one which is not connected to terminals wud get short circuited.
Now after short-circuiting that branch, its now an easy question as all are in parellel now.
So ans= C/2 + C/2 + C = 2C
I am solving these questions by nodal analysis.
Q.1- Figure-1
I have taken A to be at V potential, B to be at 0 potential and C to be at V' potential. Now in drawing-3, I have distributed same potential to the points connected by the same wires.
Now U can clearly see that except the capacitor connected to A, all other capacitors have potential difference V' across them, means all of them are in parellel. And now the circuit can be redrawn and simplified easily. And the answer is 5C/6.
Q.2- Figure-2
I have first simplified the capacitors in series. Then I have assumed A to be at 0 potential, B to be at v'' , and potentials v and v' at other points in the circuit as shown in Drawing-2.
Now in drawing -3, I have distributed the same potential to the points connected directly by a wire.
Now , it can be clearly seen that all the three capacitors of capacitance C/2 have same potential difference(v-v') across them, means they are in parellel.
Now the circuit can be redrawn as shown in drawing -4 and equivalent capacitance can be easily calculated. Ans is 3C/8.
Hope my answers are correct and U understand the method.
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
TopicsMathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
2748 Students Currently Online |
Connect with Us  |
 |
If U have any doubts in this question, ask me and u can also post ur own doubts in circuit solving here in this thread only.