In the `R_(1) = 10 Omega ,R_(2) = 20 Omega, R_(3) = 40 Omega R_(4) = 80 Omega` and `V_(A) = 5V,V_(B) = 10V,V_(c) = 20V,V_(D) = 15V` The current in the resistance `R_(1)` will be

Options

A. `0.4 A` toward O
B. `0.4A` away from O
C. `0.8A` toward O
D. `0.8A` toward O

Answer

b

Explanation

Let `V_(0)` be the potential at `O` and `I_(1),I_(2),I_(3) and I_(4)` are the current through `R_(1),R_(2),R_(3)and R_(4)` respectively Then `I_(1),I_(2),I_(3),I_(4)= 0`
`(V_(0) - V_(A))/(R_(1)) + (V_(0) - V_(B))/(R_(2)) +(V_(0) - V_(C))/(R_(3)) +(V_(0) - V_(D))/(R_(4)) = 0`
`(V_(0) - 5)/(10) + (V_(0) - 10)/(20) +(V_(0) - 20)/(40) +(V_(0) - 15)/(80) = 0`
On solving we shall get `V_(0) = 9V`
Current in the resistance `R_(1)` is
`I_(1) =(V_(0) - V_(A))/(R_(1)) = (9-5)/(10) = 0.4A`
As `V_(0) gt V_(A)`, so the currect is flowing from `O` toward A i.e. away from O.