Question

36 cells each of internal resistance 0.5 ohm and e.m.f 1.5v are used to send current to send external current through an external circuit of 2 ohm resistance the best method of grouping them is. also find the current through the external circuit.

Answer 1

Answer:

m=2,n=12

Explanation:

Total no. of cell =24

Internal resistance r =0.5π

External resistance. R =3π

best method of graping them is n/m= R/r

=> nr =mR---- eq(1 )

Total cells nm =24

n=24/m ------- eq(2)

sub. 2 in 1

24r/m =mR

24*0.5=m2*3

12= 3 m 2

m2= 4. m=2. and n =12

Answer 2

Answer:

Explanation:

here mn =36...................(1)

Internal resistance of each row =nr =0.5n

Therefore internal resistance of m rows

Is

1/r=m/nr

Or,, effective residence =nr/m=0.5n/m

For maximum current

R=effective residence

I.e 2=0.5n/m

Or. n/m=2/0.5=4....................... (2)

Multiplying 1 and 2 we get

mn *n/m=36*4

n square =144

Or. n=12 and m=3

By using formula

Maximum current =mnE/mR+nr

Here. mn=36,E=1.5v,R=2ohm,r=0.5ohm

Therefore I=36*1.5/3*2+120.5

Or I=36*1.5/6+6

=45/10

=4.5A answer