Practice problem 2

Ques) :- Find the potential difference between points F & C of the electric circuit shown in figure.

Solution :- V(F) – V(C) = - 4/3 volt.

                     Here negative sign shows that V(F) < V(C).

[ hints :- To find potential difference between any two points of a circuit you have to reach from one point to other via any path of the circuit.

                   For sloving easily Always choose a path which have least number of resistors preferably a path which has no resistance. ]

Practice Problem 1

Ques) :- Find currents in different branches of the electric circuit shown in figure.

solution:-  i 1 = 1A.
                 i 2 = 8/3A.
                 i 3 = -5/3A.

defination of junction & loop.

ques) :- what is junction?
 ans :- A junction in a circuit is a point where three or more conductors meet. It is also called nodes or branch.

  The above fig J is the junction point because at that point all the four conductors meet.


ques) :- what is loop?
 ans :-  A loop is a closed conducting path.

                         In the above fig. ABCD is closed conducting path that's we are called the above fig. is loop.     

Theory on kirchhoff's law.

Kirchhoff's laws :-

Many electric circuits cannot be reduced to simple series-parallel combinations. However, it is always possible to analyze such circuits by applying two rules. These rules are known as kirchhoff's rule.

Rules are as follows:-

1. kirchhoff's junction rule:- The algebric sum of the current into any junction is zero.

That is:-

 The junction rule is based on "conservation of electric charge" 

This law also tell:-  "The sum of all the currents directed towards a point in a circuit is equal to the sum of all the currents directed away from the point."

 applying the kirchhoff's junction rule in above fig, we find

i 1 +i 2 = i 3 + i 4.

sign convention:-      * sign is +ve if current direction towards junction .

                                    * sign is -ve if current direction away from junction.

2. Kirchhoff's loop rule:- The algebric sum of potential differences in any loop including those associated emf's  and those of resistive elements, must equal zero.

That is:-

The loop rule is based on" electrostatic field is conservative in nature".

now we study how to apply it:-

In case of only emf is given in circuit.

sign convention:- * sign is +ve when the direction of emf is from low to higher  potential.

                            *sign is -ve when the dirction of emf is from higher to lower potential.

[here lower potential means negative polarity side & higher potential means higher polarity side. The samll line of emf shows -ve polarity & the larger line of emf shows +ve polarity.]

In case when resistance and current are given.

sign convention:- *sign is positive when current is forward in direction of resistance.

                              *sign is negative when current is reverse in direction of resistance.

Now we apply both resistance & emf in a circuit & find equation for this circuit.

 The equation will be,

+ i R2 +E2 + i R1 +E1 = 0.

kirchhoff's law problem with solution.

UES_1) .Figure shows parts of a circuit. Calculate the power dissipated in 3ohm resistance. What is the potential difference   V(C) –  V(B)?

SOL :-

Applying kichhoff’s law junction law at E current in wire DE is 8A from D to E. Now further applying junction law at D. The current in 3ohm resistance will be 3  A to D.

Now the circuit is simplified.

Power dissipated in 3ohm resistance = i*i*R = 3*3*3

                                                                         =27 watt.

V(C) – V(B) : V(C) – 5*1 + 12 – 8*2 – 3 – 4*2= V(B)

                        V(C) – V(B) = 5 – 12 + 16 + 3 + 8

                                             = 20 volt.

( here <*> this symbol means multiply).