In the circuit the resistance between `A` and `C` is

Options

A. `7 r//5`
B. `r`
C. `3 r//5`
D. `r//5`

Answer

a

Explanation

Connect a battery of emf A and C let the distribution of current in the various branches of net work be According to Kirchhoff's second in a closed circuit AQBDA we have
`2rI_(1) - rI_(2) - r(I - I_(1)) = 0`
or `3 I_(1)= I+I_(2)`….(i)
in a closed circuit `BCDB`
`r(I_(1) + I_(2) - 2r(I - I_(1)- I_(2)) +I_(2) r= 0`
`3 I_(1)= 2 I- 4 I_(2)`....(ii)
from (i) and (ii) `2 I- 4 I_(2)= I+I_(2)`
or `5 I_(2) = Ior I_(2) = 1//5`

from (i) `I_(1) = (I+I_(2))//3 = [I+(1//5)]//3`
`= (6I)/(15) = (2I)/(5)`
In a closed circuit `EABCEA`
`E= 2rI_(1)+r(I_(1)+I_(2))`
or `E = 3I_(1)+I_(2) r = 3xx(2I)/(5)r+(1)/(2)r = (I)/(5)Ir`....(iii)
If `R` is the resistance of network between A and C
Then `E = IR`....(i)
Alternate method
Here `R_(1)= r : R_(2) = 2r`
The effective resistance of net work between `A` and `C` is
`= (R_(1)(R_(1) + 3R_(2)))/(R_(2)+3R_(1)) = (r(r+3 xx 2r))/(2r+3r) = (7r)/(5)`