Complete notes of current electricity –

Physics classes by Nayan jha providing Complete notes of current electricity . As you know i am already provided the notes of all topics related to current electricity. But for the students to get the Complete notes of current electricity , here students can find each topics of current electricity at one page .

1. Electric current,

2. Drift velocity , mobility  , relation between drift velocity and current

3. proof of Ohm’s law on the basis of drift velocity .

4. Current density , resistance , conductance , resistivity, conductivity , relation between current density and drift velocity.

4. Effect of temperature on resistance and resistivity  , graph of resistance(resistivity) Vs temperature for conductors/alloy /semiconductors .

5. Ohmic and non-ohmic conductors ( graph) ,

6.Carbon resister and color coding .

7. Combination of resistors ( series and parallel ) .

8. Internal resistance (r) , terminal potential (V)  and emf (E)  of the cell and relation between them . also graph between ( V,E, r and R)  .

9.Combination of cells (series, parallel ,and mixed grouping) .

10. Kirchhoff’s law (junction and loop rules) ,

11.Wheat -stone bridge .

12. Meter bridge.

13. Potentiometer and its applications ( effects on null points on different -different conditions ).

14. Electrical energy and power .

 

Potentiometer

In this topic we will discuss about Potentiometer,its working principle and its applications. Also we will know sensitivity of potentiometer.

Before to know about this topic student must read the following topics –

i.- Kirchhoff’s law,to read this topic click here-

ii.- Wheatstone bridge,to read this topic click here-

iii.-Mete bridge,to read this topic click here-

Potentiometer – An instrument used to measure accurately the EMF or potential difference is called potentiometer.

Principle– Its working based on the principle that the potential drop across a conductor is directly proportional to the length of the conductor , if constant current passes through the conductor having uniform area of cross-section .

Construction –

 As we know ,

potential V=IR

V = (ρl/A) I

Since current is constant and area of cross-section is uniform  so V α l

Or V = K l

K = V/l ( K is a constant called potential gradient)

[ calculation of potential gradient ( students don’t have to derive it in case of potentiometer it is useful to solve  the numerical questions) – if Rh = r and resistance of the wire is r and E is the emf of  driver cell  then ,  current through the circuit is , i = E/ (R+r)

There fore potential difference at the end  the wire is v= i r

V= {E/ (R+r) } r . so potential gradient  K = V/l = {E/ (R+r) } r/l. ]

 

Application of potentiometer –

1. Determination of potential difference using potentiometer –

All the arrangements are shown in the figure-

Let key K is closed then current i starts flowing through the circuit , if  R_AB is the resistance of the wire then , We can write current i = ε/R+R_AB ;

Now we close the key K₁ also , in that case potential across the R₁ is ‘V’ and null point J is at length ‘l’ as shown in figure .

Then V = Kl ;

But , K = V_AB/ L   so we can write V =( V_AB/ L)l   [L Is the total length of potentiometer wire]

Here V_AB = i R_AB = (ε/R+R_AB)R_AB ;

Therefore  V = [ {ε/(R+R_AB)}l R_AB]/L ;

2. Comparison of emf of two cells –

A battery of emf ε is connected between the terminals A and B with variable resistance R_h , Now we have to compare the emf  ε₁ and ε₂ . We arrange the two cells having emf ε₁ and ε₂ as shown in figure . At first we connect the terminals 1 and 3 , let at that time J₁ is the position where we grt null point , let length of A J₁ is l₁ so we can write  ε₁= Kl₁ ……….(i)

Again 1 and 3 is de-attached and terminal 2 and 3 is connected  now this time we get null point at J₂ , let AJ_2­ = l₂ . then we can write   ε₂= Kl₂ ………………….(ii)

Dividing equation (i) by equation (ii) we get  ,  ε_1/ ε₂= l₁/l₂ …………..req. eq.

Precautions of experiment-  (i) Current through the circuit must be constant .

(ii) the emf of the driving cell (ε) must be greater than emf of the cells whose value has to be compared .

3. Determination of the internal resistance of the cell –

To find the internal resistance r  of the cell  emf ‘ε’ we setup the circuit as shown in figure.

Initially we close the switch ‘K’ and switch K₂ is remain open .

In that case null point is at J₁ , let AJ₁=l₁

Then we can write ε=Kl₁………………(i)

Now the switch K₁ is closed then the potential across the resistance R is   ‘V’ . And in that case we get the null point at J₂ . And AJ₂=l₂

Then we can write V = Kl₂   ………….(ii)

Dividing (i) by (ii) we get  ε/V = l₁/l₂ ……………(iii)

But we know that internal resistance of the cell   r = (ε-V /V)R = [ε/V-1]R ……….(iv)

And from equation (iii) and (iv) we get , r = [l₁/l₂ -1]R ………Req. Eq.

Sensitivity of potentiometer –

The sensitiveness of the potentiometer means the smallest potential difference which can be measured with the help of potentiometer .

The sensitiveness of the potentiometer can be increased ;

(i)by increasing the length of the wire of the potentiometer ,

(ii) by decreasing the value of potential gradient using reducing the current by increasing the resistance .

Difference between EMF of a cell and potential difference –

(i)  EMF -It is related with source of current

P.D. – It is related with any two points of the closed circuit.

(ii)  EMF – It cause of potential difference between two point of circuit .

P.D. – It is an effect .

(III)  EMF – It is the amount of work done in taking a unit charge once round the complete circuit .

P.D. – It is the amount of work done to bring unit charge from one point to another point .

(iv)  EMF – It is equal to sum of potential difference across all components of the circuit ( Including all          the cells connected  in the circuit .

P.D. – In a closed circuit the different components of the circuit have different potential difference .

(v)   EMF – It is independent of the external resistance and current drawn from the circuit .

P.D. – It depends on the resistance between two points of circuit and current flowing through it .

Meter bridge or slide wire bridge

In this topic we will discuss about Meter bridge or slide wire bridge . Principle of Meter bridge or slide wire bridge  ,  its working and, end errors with correction of end error.

Before to know this topic students must read the following topics.

i. Kirchhoff’s rule , To get the notes of this topic click here-

ii. Wheatstone bridge , To get the notes of this topic click here-

Meter bridge or slide wire bridge –

It is a device used to find the value of unknown resistance . It works on the principle of wheat-stone bridge .

It consist a wire of Constantan or Manganin of one meter length and having uniform area of cross-section . As shown in figure wire is stretched between AC , a cell with a key ‘K’ a resistance box R( known resistance ) is attached and an unknown resistance S  , in two gap as shown in figure .

When switch ‘k’ is on , we start sliding the jockey from A to c , we notice there is a point where galvanometer shows no deflection ( known as null point) . Let this point is at B , let  AB = l

so ,      BC = 100-l

here R_AB=ƍl/A ( Where A is the area of cross-section)

R_BC = ƍ(100-l)/A

According to wheat-stone bridge principle (P/Q=R/S) We can write

If meter bridge is balanced then R/R_AB = S/R_BC

Or , R/ ( ƍl/A) = S / ƍ(100-l)/A

Or, R/l =S/(100-l)    so    S = R(100-l)/l ……………eq.

 

To see the video of meter bridge click on the link given below-

 

Error end in the meter bridge- cause of the zero end error is following-

(a) The resistance of the copper strip and connecting wire is not taken in account .

(b)The zero mark of the scale may not start from the position where copper strip leaves the wire and 100 cm  mark may not on the position where the bridge wire just touches the copper strip .

Processes to remove the end error –

By changing the ‘R’ with ‘S’ and ‘S’ with ‘R’ and repeating the experiment . The average value of S determined earlier and now is free from end error .

*** NOTES-  1. Null point ( where galvanometer shows no deflection) should be at the middle of the bridge wire . Because it becomes more sensitive when all four resistances are nearly same.

2.The meter bridge can’t used to measure very high or very low resistance .

3. the balanced point not effected after interchanging the position of galvanometer and battery .

Wheatstone bridge

In this topic we will discuss about Wheatstone bridge ,its principle and balanced condition for Wheatstone bridge .And some important points for Wheatstone bridge.

Before to know about Wheatstone bridge , students must read these topics first-

i. Kirchhoff’s rule , to get the notes on this topic click here-

ii. Ohm’s law, to get the notes on this topic click here-

iii. Emf of the cells, To get the notes on this topic click here-

Wheatstone bridge –

It is an arrangement in which four resistances are arranged in the form of a bridge which is used for measuring one unknown resistance in terms of other three resistances .

Wheatstone bridge principle – It’s principle states that when four resistances P,Q,R and S are connected with a galvanometer and a cell in the form of bridge as shown in figure , the bridge is said to be balanced when there is no deflection shown by the galvanometer for this condition P/Q = R/S .

As shown in figure , resistance P is in arm AB , Q is in arm BC , R is in arm AD and S is in arm DC . A cell of emf ‘ε’ and galvanometer is connected as shown in figure with  keys K₁ and K₂ . On closing key K₁ and then key K₂ if galvanometer shows no deflection then it is said that bridge is balance , and we will get  P/Q = R/S  .

As shown in figure I is the current flowing through the cell , when it reaches to point A it divided in two parts I₁ towards AB and ( I-I₁ ) towards AD . Again when I₁ reaches to B then current I_g flows through galvanometer and ( I₁-I_g) towards C and current I-I₁+I_g towards DC .

Now applying Kirchhoff’s loop law in loop ABDA we get,

I₁P+I_gG-(I-I₁)R=0 …………………..(I)

Again from loop BCDB we get   (I₁-I_g)Q – ( I-I₁+I_g)S-I_gG=0………………(II)

If galvanometer shows no deflection i.e. I_g=0 then eq (ii) may be written as

I₁P  =  (I-I₁) R ………………….(III)

And eq. (ii) may be written as ,

I₁Q  =  (I-I₁) S ……………………..(IV)

Dividing (iii) by (iv) we get P/Q =R/S  This proves wheat-stone bridge .

Two view the video on Wheatstone bridge click here-

 

*****Important point for wheat-stone bridge ***

(1) Wheat-stone bridge is a null method , hence the measurement of unknown resistance is very accurate.

(2) Wheat-stone bridge remain unaffected by the internal resistance of the cell and galvanometer.

(3) it is not used to measure very high or very low resistances .

(4) It becomes very sensitive when  resistance of all four arms are nearly same .

 

Kirchhoff’s rules

In this topic we will discuss about , Kirchhoff’s rule and explanation of  its type.And also we will discuss, how Kirchhoff’s second rule loop rule/ potential rule , support the law of conservation of energy ?

Kirchhoff’s rule 

First rule ( junction rule /current rule ) –  According to this law when current flowing through several wires meet at a point ( junction) the at a junction in closed circuit is always zero .

[ sign convention –  generally when current flows towards the junction is taken positive and when it moves away from the junction is taken negative . ]

As shown in figure,  ‘o’ is the junction then according to Kirchhoff’s law  .

-I₁+I₂-I₃+I₄+I₅ = 0 …………………..Eq.

Then from the above relation we can say at a junction current entering is equal to current leaving .

Second rule ( loop rule / voltage rule ) – According to this rule for a closed loop the algebraic sum of change in potential must be zero . In another words we can say algebraic sum of potential differences across the cells and product of current and resistance across the resistor should be zero .

[ ***sign convention-  select any point on the closed circuital loop you can move along your desired direction (clock or anti-clock wise direction) , when you reaches to the resistor then see the your moving direction and direction in which current flowing through the circuit the sign of ‘IR’( product of current and resistance ) if it is same the sign will be positive , if it is opposite then sign will be negative . And sign of potential or EMF of the cell will be positive if you touch positive terminal first , and sign will be negative if you touch negative terminal first ].

To understand the loop rule see the following example –

Suppose we have a loop shown in fig;-

from the loop ABCDEFA,     -ε₂+ ε₁+i₂R₂ -i₁R₁=0

or , ε₁– ε₂= i₁R₁– i₂R_{2 ……………………..(i)}

again from the loop BEDCB ,  ε₂-i₃R₃-i₂R₂=0

or,  ε₂= i₃R₃+i₂R_{2………………………..(ii)}

and from the loop ABEFA, ε₁ -i₃R₃-i₁R₁=0

or, ε₁ =i₃R₃+i₁R_{1………………………….(iii)}

using equations (i),(ii),and (iii) we can find the all unknown quantities.

*** Kirchhoff’s second rule loop rule/ potential rule , support the law of conservation of energy –

In a closed loop the charges calculated around a closed loop in a particular direction due to electrostatic force which is conservative in nature . The work done by the electrostatic force on the circulating charge in a closed loop is zero . Due to it , the net charge in the energy of charge when goes on a closed loop is zero. It means there is no gain in energy  by circulating charges around a closed loop in a particular direction which is law of conservation of energy . Thus Kirchhoff’s supports the law of conservation of energy . ***

Grouping of cells

In this topic we will discuss about grouping of cells in series and parallel. Also we will discuss about the mixed grouping of cells .

Before to know about this topic student must know about emf of cells, internal resistance also the relation between emf / terminal potential and internal resistances , and the grouping of resistances .

To know about emf of cells click here-

To know about the relation between emf / terminal potential and internal resistances , click here-

To know about grouping of cells click here-

Combination of cells – 

(1) When two cells are connected in series  –  When negative terminal of first cell is connected to positive terminal of second then it is said that cells are connected in series . let the EMF of the cells and their internal resistances are [ ε₁ , r_1- ] and [ε₂, r₂ ] , are connected with external resistance R .

Let V_AB and V_BC are the potential differences across AB and BC respectively . since the combination is series then current will be same (let I )

Then V_AB = ε₁ – Ir₁ ……………..(I)

And V_BC =  ε₂ -Ir₂  …………………(ii)

Then V_AC = V_AB – V_BC = (ε₁ – Ir₁ )– (ε₂ -Ir₂  ) = ε₁+ ε₂ – I(r₁ + r₂ )  …………(iii)

Let ε_eq is the equivalent resistance of the combination r_eq is the equivalent internal resistance of the combination .

Then we can write , V_AC =  ε_eq – Ir_eq ……………(iv)

On comparing eq. (iii) and (iv) we get ,

ε_eq= ε₁+ ε₂  and r_eq = r₁ +r₂

** I f there are n cells connected in series then we can say

,    ε_eq = ε₁+ ε₂ + ε₃+ ε₄………… ε_n      and  , r_eq = r₁ +r₂ + r₃ +r₄……….. r_n    ;

[ ***special cases –

Case 1 –  If the direction of the combination of  cells are reversed then

, ε_eq = ε₁– ε₂ and   r_eq = r₁ +r₂

Case 2 – If n identical resistances are connected in series then

ε_eq = n ε   and  , r_eq = nr and hence,  current through the circuit  I = n ε   /( nr+R)     ].

 

(2) When two cells are connected in parallel –  If positive terminals of the cells are connected at one point and all negative terminal at another point then such kind of combination of cells are said to be in parallel combination .

Let the EMF of the cells and their internal resistances are [ ε₁ , r_1- ] and [ε₂, r₂ ] , are connected with external resistance R .

Let I₁  and I₂ are the current flowing  from both the cells as shown in figure , V is the terminal potential . then I ₁ =( ε₁-V)/r₁   I₂ = (ε₂-V)/r₂

Net current I =  I₁ + I₂  ;

I =  (ε₁-V)/r₁  +  (ε₂-V)/r₂…………………..(i)

If  ε_eq is the equivalent resistance of the combination r_eq is the equivalent internal resistance of the combination .

Then I = (ε_eq-V)/r_eq ………………….(ii) ,

From equation (i) and (ii)

(ε_eq-V)/r_eq = (ε₁-V)/r₁  +  (ε₂-V)/r₂ ;

(ε_eq-V)/r_eq =  [ ε₁/r₁  +  ε₂/r₂] -V [1/r₁ + 1/r₂ ]

(ε_eq-V)/r_eq  = (ε₁r₂  +  ε₂r₁)/r₁r₂   – V [1/r₁ + 1/r₂ ]

Then V = (ε₁r₂  +  ε₂r₁)/r₁+r₂ – I r₁r₂/r₁+r₂ ………………………(iii)

But we know V = V_AB = ε_eq – Ir_eq ………………(iv)

On comparing (iii) and (iv) we get ,

ε_eq = (ε₁r₂  +  ε₂r₁)/r₁+r₂

and , r_eq = r₁r₂/r₁+r₂ or [ 1/r_eq= 1/r₁ +1/r₂ ].

[***If there are n cells connected in parallel then ;ε_eq /r_eq  = ε₁ /r₁  + ε₂ /r₂  + ε₃ /r₃  +………….ε_n /r_n  and,   1/r_eq = 1/r₁ + 1/r₂+1/r₃………1/r_n].

[****Special case –

If n identical cells are connected in parallel then  ε_eq =  ε

Then current in the circuit will be given as  I =  ε /R+ r/n  =    n ε /nR+r ;     ]

 

(3) Mixed grouping of cells – L et n identical cells are connected in series and m such series combination in parallel as shown in figure .

 

As shown in each row there are n cells in series then total emf and resistance of each row will be ‘nε’ and ‘nr’ respectively   . Since there are ‘m’ rows in parallel then equivalent resistance will be given as

1/r_p =  1/nr+1/nr+1/nr………….to m terms ;

So, r_p = nr/m ,

Therefore total resistance of the circuit will be ;- R+( nr/m) ;

Current through the circuit flows will be –  I = nε / R+( nr/m) = mnε/MR+nr ; current

[ ** special case – Through the circuit will be maximum when internal resistance is equal to external resistance   i.e. mR = nr ;or , R = nr/m   ].

Relation between electromotive force,internal resistance and terminal potential.

In this topic we will discuss about electromotive force,internal resistance and terminal potential,relation between them . Also we will plot the graph between them .

EMF , Internal resistance and terminal potential difference of a cell –

EMF of a cell –  As we know electric current flows through a complete closed circuit ,when there is an source of external force which can bring the ions from low potential to high potential . This external for which makes the current carriers to move in a definite direction called electromotive force (EMF) . In another words we can define ‘ It is the maximum potential difference between two electrodes of the cell when no current flows through the circuits i.e. in open (off) circuit‘ .

Unit of the EMF is Volt (V) and it is generally represented by ‘ε’.

Internal resistance –

 It is the resistance offered by the electrodes and electrolytes of a cell to the passes of electric current through it. It is denoted by ‘r’ .

its value is generally very low its value increases when cell becomes older and older .

The value of internal resistance depends on (a)area of the electrodes immersed in the electrolytes ,(ii)nature of electrodes , (iii) distance between the electrodes and (iv)nature of electrolytes .

Terminal potential difference –  Terminal potential difference of a cell is defined as the potential difference between the two electrodes of a cell in a closed circuit . it is denoted by ‘V’ . or simply we can define when current flows through a circuit then the potential difference across the end of the resistor. In a closed circuit the terminal potential difference of a cell is always less than the EMF of the cell .

Suppose , ε is the EMF of the cell , r is its internal resistance  which is connected with an external resistance R with the help of a key ( K ) as shown in the figure ,

If the key is not closed , the circuit remain open no current flows through the circuit , the  reading of the voltmeter is equal to emf ε . [ i.e. terminal potential of the cell is equal to the emf of the cell( V= ε) ] .

When key is closed then current start flowing through the circuit,

in that case emf of the cell   ε= I(R+r) = IR +Ir = V +Ir

so we can write , V = ε -Ir ………………..eq.

from the above equation  we just come to know ;

case (1)- if circuit is open ( off) I=0   so, V = ε

case (2) – If circuit is closed (on) , then  V < ε

case (3) –  ** During the charging V > ε ;

 

From the above equation  V = ε -Ir

We can find  internal resistance  r = (ε -V)/I

But I=v/R  so we can write   r = ( ε-V/V) R ……eq. to find the internal resistance of the cell .

Graph between (a) ε and R (b) V and R (c) V and I; are  given below –

 

Combination of resistances

In this topic we will discuss about combination of resistances , i.e series and parallel combination of resistances .

Combination of resistances –

Series combination –  In series combination resistors are connected end to end with one another . resistors are said to be connected in series , when same current passes through the resistors  when potential difference applied across the connected  resistors .

Let , three resistors  of resistance R₁, R₂,and R₃ are connected in series with external source E as shown in figure (a) . Let  I be the current flowing through the each resistors , their respective potential differences are V₁, V₂ and  V₃ .

According to Ohm’s law  here,  V₁ = I R₁  ,    V₂ = I R₂   ,  and   V₃ = I R₃

Let, R_S is the equivalent resistance of the combination as shown in figure (b) ,

then , V = I R_S

But net potential V = V₁+ V₂ +V₃

Then  IR_S  = I R₁ + I R₂ + I R₃

So ,  R_S  =  R₁ +  R₂ +  R₃  ……………………This the equation of equivalent resistance .

 Parallel combination –  Two or more resistors are said to be in parallel if potential difference across each resistors are same but currents are different . In parallel combination one end of each resistors are connected at one point and other ends at another point .

Let three resistors of resistance R₁ , R₂ and R₃ are connected in parallel as shown in figure (a) with external source . let V be the potential difference of the end A and B . I₁, I₂ and  I₃ are the currents flowing through the resistors .

 

According to Ohm’s law  I=V/R .

SO, I₁=V/R₁ , I₂=V/R₂ and I₃=V/R₃

Let, R_p is the equivalent resistance of the combination as shown in figure (b)

Then I=V/R_p

But I = I₁+ I₂ + I₃

V/R_p = V/R₁ + V/R₂ + V/R₃

SO, 1/R_p = 1/R₁ + 1/R₂ + 1/R₃ …………………….. equation for equivalent resistance .