# Equivalent resistance?

Mar 17, 2018

$C$

#### 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 $\frac{7}{12} \Omega$

So, (R×1)/(R+1) =7/12

Or, $12 R = 7 R + 7$

So, $5 R = 7$

Or, $R = \frac{7}{5} \Omega$