Question

Equivalent resistance?

Answer 1

`C`

Explanation:

You can think this way,

Let,equivalent resistance of the whole circuit between point `A` and `B` is `R` when, the link is not present.

If the link of `1 Omega` was present,it would have in parallel to `R`

So,equivalent resistance between `A` and `B` would have been `(R×1)/(R+1)`

Which is given to be `7/12 Omega`

So, `(R×1)/(R+1) =7/12`

Or, `12R =7R +7`

So, `5R=7`

Or, `R=7/5 Omega`