A network of resistance is constructed with `R_1 and R_2` as shown in fig. The potential at the points 1,2,3 …., N are `V_1, V_2,V_3,…,V_n,` respectively, each having a potential k times smaller then the previous one.

The current that passes through the resistance `R_2` nearest to the `V_0` is

Options

A. `((k-1)^(2))/(k) (V_(0))/(R_(3))`
B. `((k+1)^(2))/(k) =(v_(0))/(R_(3))`
C. `(k+(1)/(k^(2)))(V_(0))/(R_(3))`
D. `((k-1)/(k^(2))) (V_(0))/(R_(3))`

Answer

D

Explanation

Given `V_(1) = (V_(0))/(k), V_(2) = (V_(1))/(k), V_(3) = (V_(2))/(k), I = I_(1) +I_(2)`

`(V_(0)-V_(1))/(R_(1)) = (V_(1)-V_(2))/(R_(1)) +(V_(1) -0)/(R_(2))`
`(V_(0) - V_(1)//k)/(R_(1)) = (V_(0)//k-V_(0)//k^(2))/(R_(1)) +(V_(0)//k)/(R_(2))`
`(R_(1))/(R_(2)) = ((k-1)^(2))/(k)`
current in `R_(1)` and `R_(3)` will be the same:
`(V_(n-1) - V_(n))/(R_(1)) = (V_(n))/(R_(3))`
or `(V_(n-1)-(V_(n-1))/(k))/(R_(1))=(V_(n-1))/(kR_(3))`
or `R_(1) = R_(3) (k-1)`
put the value of `R_(1)` in the above expression, we get
`(R_(2))/(R_(3)) = (k)/(k-1)`
Current in `R_(2)` nearest to `V_(0)` is
`I_(2)=(V_(1))/(R_(2))=(V_(0)//k)/(R_(3)((k)/(k-1)))=((k-1)/(K^(2)))(V_(0))/(R_(3))`