Current & Electricity

Relation b/w Electric current & Drift velocity :-

Let “E” be the electric field applied across a conductor of length “l” & cross sectional area “A” .Suppose the number of free electrons per unit volume of the conductor is “n” .

            Then total number of electron in the conductor  = n (A .l)

             Now , total charge  within the conductor

                        q  =   total number of electrons  X   charge of each electron

                        q =   n A l  X e

            or         q  =   n e A l   

 therefore time taken by the electron to cover the length “l” is

                                 length of the conductor

t  = ---------------------------------

               drift velocity

             l

t = ---------------

            v_d

    

Now the current flowing through the conductor is

              q                    n e A l

I   =   -------------  =  --------------- = n e A v_d

               t                       l/v_d

hence 

                                    I = n e A v_d

 

Resistivity or Specific resistivity :- From the relation   

                                           

                                                             l

                                            R =  ρ----------------                                            

                                                              A

                                           R  X  A

We have resistivity  ρ = -----------------

                                                l

If  “l”   = 1 unit , A = 1 unit

Then        ρ  = R 

Hence resistivity is numerically equal to the resistence of a conductor having unit length & unit area of cross section .

It is the property of the material of the conductor .

Its S.I. unit is  Ohm-meter (Ω-m).

Conductivity(σ ) :-  It is defined as the reciprocal of resistivity . So

                                        1

Conductivity σ =---------------------------

                                  Resistivity 

Its S.I unit is   1/ Ohm –meter  =  (Ω-m)^-1

 

Temperature dependence of Resistivity

The resistivity of conductor is increased with temperature , while in case semiconductors & insulators ,the resistivity decreases with increasing temperature

Colour code for carbon resistors:-Carbon resistors have four bands or rings of different colours.the first two bands indicates the decimal multiplier. The fourth and last band indicates the tolerance.

The following table gives the colour code for carbon resistors 

Colour	Number	Multiplier
Black	0	100or 1
Brown	1	101
Red	2	102
Orange	3	103
Yellow	4	104
Green	5	105
Blue	6	106
Violet	7	107
Gray	8	108
white	9	109

Tolerance (%):

GOLD ± 5%

SILVER ± 10%

NO COLOR  ±20%

EX:-If the four colors are yellow ,orange ,blue & gold, the resistance value will be : 43X10⁶ ± 5%.

 

Kirchhoff’s rule ---First rule or Junction rule  It states that the algebraic sum of all the junction is Zero .

Ex- The current entering the junction are taken as positive  & the current leaving the junction are taken as negative . 

From the given figure applying kirchhoff’s rule we get

I₁ -  I₂  +    I₃  +     I₄  -     I₅   +      I₆   =   0

 

Kirchhoff’s IInd law or loop law ----

The algebraic sum of changes in potential around any closed loop involving resistors & cells in the loop is zero .

Sign convention –

(i)                The e.m.f. of a cell is taken negative if one moves in the direction of increasing potential (i.e. from –ve pole to + ve pole ) through the cell & is taken positive if one moves in the direction of decreasing potential ( from +ve pole to –ve pole) trough the cell . It means while traversing a loop  if –ve pole of the cell is incountered first then its e.m.f. is negative otherwise positive .

(ii)             The product of resistence & current in an arm of the circuit is taken +ve  if the direction of the current in that arm is in the same sense as one moves in a closed path & taken –ve if the direction of current in that arm is opposite to the sense as one moves in a closed   path .

(iii)          

     We can traverse in aclosed path of a circuit in aclockwise or anticlockwise direction.

Ex- Applying kirchoff’s IInd rule in aclosed loop ABEFA

     I₃ R ₂          +  I₁ R ₁           - E₁    =  0

             E₁    =     I₃ R ₂          +  I₁ R ₁          

Similarly  for closed loop BCDEB -

        I₃ R ₂          +  I₂ R 3           -  E₂    =  0

     

         E₂    =  I₃ R ₂          +  I₂ R 3          

 

Wheat stone bridge – it is an application of kirchhoff’s law .

Wheat stone bridge consists of four resistance  P,Q,R &S connected in a different arms of a quadrilateral ABCD . cA source of EMF  & key is connected along one of the diagonal & a galvanometer is connected along another diagonal .

 In the case of  a balanced bridge , there is no current through the galavanometer  i.e. I_g   = 0

Then                             P/Q =R/S

Proof –

Applying kirchhoff’s  IInd  law in the closed loop ABDA

         I₁ P  - (I-I₁ ) R = 0

                 I₁ P   = (I-I₁ ) R ------------------------(1)

Applying kirchhoff’s IInd law in the closed loop BCDB

  

               I₁ Q  - (I-I₁ ) S = 0

                I₁ P           =     (I-I₁ ) R  ---------------(2)

Dividing eq (1) by (2) –

       I₁ P                         (I-I₁ ) R

----------------      = --------------------

       I₁ Q                         (I-I₁ ) S

    P/Q   =   R/S                                hp 

 

QUESTION:- In the fig different resistors are connected. Find the net current drawn from the cell.

SOLUTION:- In the given diagram

                   According to the balanced condition of  WHEAT STONE BRIDGE

P/Q =R/S

2/4 = 3/6 =1/2

So, the bridge is balanced & there will be no current through the 5Ω resistor.

To find the equivalent resistance between A& C:-

                                                  Since 2Ω &4Ω are connected in series.

                                                                 R_ABC = R_AB + R_BC

                       = 2Ω + 4Ω

              = 6Ω 

Again 3Ω & 6Ω are connected in series

 R_ADC =  R_AD + R_DC

        = 3Ω + 6Ω

= 9Ω

Now these resistances of 6Ω & 9Ω are connected in parallel. So, the equivalent resistances between A& C is

1/ R_AC = 1/6+1/9

      = 18/5

       =3•6Ω

So,the equivalent resistance of the circuit is

R = 3•6Ω + 2∙4Ω  = 6Ω

Hence, the current drawn from the cell is

I = 12Ω/6Ω

= 2Amp

 

METER BRIDGE OR SLIDE WIRE BRIDGE :- It is the practical application of Wheat stone bridge. It consist of a wire of length 1m & uniform cross-sectional area.

From the condition of balanced Wheat stone bridge

   R/S = rl₁/r(100-l₁)

R/S = l₁/(100-l₁)

Where ‘r’ is the resistance of wire per unit length of meter bridge.

S = R (100-l₁)/ l₁

Using this formula, we can find out the unknown resistance.

POTENTIOMETER :- it is an instrument which is used to measure potential difference .

PRINCIPLE OF POTENTIOMETER WIRE :- If a constant current is passing through the wire of uniform area of cross-section of potentiometer, then the potential drop across any portion of the wire is directly proportional to the length  of that portion of the wire.

We know that,

                      V = IR

                  But R = ρl/A

V =Iρl/A

Or V =kl

Where k = Iρ/A is called the potential gradient.

V

APPLICATION OF  POTIOMETER WIRE :-

1.To find out ratio of emf of two cells:- there are two cells having enf E₁ & E₂ connected as per circuit diagram. First we use the cell of emf E₁ & suppose by using this cell, the null point comes out at a length l₁ from  end A & by using the second cell of                                         

emf E₂ suppose the balancing length is        

at length `l₂’ from end A.   

                                      

           Then E₁=kl₁ &

      E_{2 =}kl₂

      Therefore       E₁/ E₂=l₁/l₂

2. To find the internal resistance of cell:-

                   In the given circuit diagram , a cell of

                   emf  E is connected alongwith a key

                   & resistance box. Suppose the internal 

                          

   resistance of the cell is `r’.

  

  let when the key K₂ is open , balance point isat length l₁ then

          E=kl₁       ………………………. (1)

            Now when the key k₂ is closed , suppose the Balance point is at length l₂   then terminal potential difference of cell is

            V=kl₂    ………………….. (2)

  Therefore      E/V=l₁/l₂

 But we have the relation

                           r=R(E/V -1)

    therefore    r=R(l₁/l₂ -1)

               by knowing the length l₁ & l₂  , we can find out the value of internal resistance `r’

 

SOME IMPORTANT QUESTION:-

1. Derive relation between current & drift velocity           (CBSE 2003) 

2. Define term resistivity & conductivity & state their units.

                                                                                       (CBSE 2001-03)

3. Explain colour code for carbon resisters with illustration.

4. The variation of potential difference

     V with length l in case of two potentiometer      

         

      X & Y is as shown in fig. Which one of

      these two will you prefer for comparing

      emf ‘s of two cells & why? (CBSE 2006)

5. State Wheat stone bridge principle & deduce it using Kirchhoff’s                                        laws.  (CBSE 2002)

6. With a circuit diagram, explain how a meter bridge can be used to     find the unknown resistance of a given wire. State the formula used.      

                                                                            (CBSE 2001)

7. State principle of Wheat stone bridge. Draw a circuit diagram used

    to compare the emf of two cells. Write the formula used. How the

    sensitivity of a potentiometer can be increased.    (CBSE 2006)

8. Copper wire of resistance R₀ is stretched till its length is increase  to

    `n’ times of its original length. What is its new resistance.

9. A 10meter long wire AB of uniform area                                      

of cross-section & 20Ω resistance is used as

 a potentiometer wire. This wire is connected 

                   

in series with a battery of 5V & a resistor of

480 Ω. A unknown emf is balanced at 600cm

of the wire as shown in fig. Calculate-

   1-The potential gradient for the potentiometer wire.

   2-The value of unknown emf

                                                                (CBSE 1998)