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. ]

Capacitance of a spherical conductor.

     v Capacitance of a spherical conductor.

 When a +ve charge q is given to a spherical conductor of radius R, the potential on it is, 

     From the above expression we find that,

                                               

Thus, capacitance of the spherical conductor is,

From this expression we can reach at the following conclusion. 

(1).

  

Which we have directly stated that C depends on the dimensions of the conductor.

If two conductors have radii R 1 & R 2 then,

(2).Earth is also a spherical conductor of radius 

 

m.    The capacity of earth is therefore,

                                             

C = 711  microF                       [ F= farad].

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.