Grouping of cells

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 , r1- ] and [ε2, r2 ] , are connected with external resistance R .

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

Then VAB = ε1 – Ir1 ……………..(I)

And VBC =  ε2 -Ir2  …………………(ii)

Then VAC = VAB – VBC = (ε1 – Ir1 )– (ε2 -Ir2  ) = ε1+ ε2 – I(r1 + r2 )  …………(iii)

Let εeq is the equivalent resistance of the combination req is the equivalent internal resistance of the combination .

Then we can write , VAC =  εeq – Ireq ……………(iv)

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

εeq= ε1+ ε2  and req = r1 +r2

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

,    εeq = ε1+ ε2 + ε3+ ε4………… ε    and  , req = r1 +r2 + r3 +r4……….. rn    ;

[ ***special cases –

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

, εeq = ε1– ε2 and   req = r1 +r2

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

εeq = n ε   and  , req = 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 , r1- ] 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  εeq is the equivalent resistance of the combination req is the equivalent internal resistance of the combination .

Then I = (εeq-V)/req ………………….(ii) ,

From equation (i) and (ii)

eq-V)/req = (ε1-V)/r1  +  (ε2-V)/r2 ;

eq-V)/req =  [ ε1/r1  +  ε2/r2] -V [1/r1 + 1/r2 ]

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

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

But we know V = VAB = εeq – Ireq ………………(iv)

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

εeq = (ε1r2  +  ε2r1)/r1+r2

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

[***If there are n cells connected in parallel then ;εeq /req  = ε1 /r1  + ε2 /r2  + ε3 /r3  +………….εn /rn  and,   1/req = 1/r1 + 1/r2+1/r3………1/rn].

[****Special case –

If n identical cells are connected in parallel then  εeq =  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/rp =  1/nr+1/nr+1/nr………….to m terms ;

So, rp = 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 caseThrough the circuit will be maximum when internal resistance is equal to external resistance   i.e. mR = nr ;or , R = nr/m   ].

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