Question

solve if any one has potential i will give max point and brainliest.


all three questions must be answered

Answer 1

26.
I = m e / m R

as we know this formula works when r <<R

I = e / R
I = 2 /6
I= 0.33 A
ans (1)

27..
see in pic attached

28.

Eq
as the three are connected in wrong order then they are canceled which are 6
and six left

Eeq = n x E.mF
E= 6 x2
E = 12

Answer 2

all internal resistance connected in parallel . after derrive we get the formula ,
I = ( E1/r1 + E2/r2 + E3/r3 + E4/r4)/( 1 + R( 1/r1 + 1/r2 + 1/r3 + 1/r4 )

where , E1 , E2, E3 and E4 are potential difference between internal resistance r1 , r2 , r 3 , r 4 and R is the external resistance .

so,
I = ( 2/1 + 2/1 + 2/1 + 2/1 )/( 1 + 6( 1/1 + 1/1 + 1/1 + 1/1) A

= ( 2 +2 + 2 + 2 )/( 1 + 6 × 4)

= 8/25 A

= 32/100 A

= 0.32 A
option (1) is correct .
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27 )) General formula for solving this type of question is :::::::::
Let their are n cell each of EMF " E " and internal resistance " r " are connected in series , x of them are wrongly connected then ,

I = ( n -2x )E/( R + nr)
where R is external resistance .

now, here n = 4
x = 1
E = 1 V
r = 0.5 ohm
R = 1 ohm

so,
I = ( 4 - 2 × 1)× 1 /( 1 + 4× 0.5 )

= 2/3 A
option (2) is correct .

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28))
effective emf = ( n -2x)E

where n is no of cells , x is no of cells are connected wrongly . E is EMF of each cell .

here,
n = 12
x = 3
E = 2V

so, effective e.m.f = ( 12 - 2×3)× 2 = 12 V
so, option (2) is correct .