Question

The equivalent resistance across AB is:

Answer 1

We connect the current network together. Here the endpoints form a balanced Wheatstone bridge. Therefore by considering the corresponding resistances between the bridge and removing the conducting wire in between considering the current in the circuit turns to zero when a galvanometer is placed. Hence we can find the equivalent resistance between $A$ and$B$.

Complete step by step solution:
Consider the figure where we place the points $F,{\text{ }}E,{\text{ }}H,{\text{ }}G,{\text{ }}C,{\text{ }}D,$ at the ends of the bridge and the center conducting wire $GH$, applying the concept of Wheatstone bridge we get,

Let us take $CDEF$ as a balanced Wheatstone bridge. Because each arm resistance will be equal to $2\Omega$ so the equivalent resistance for the bridge between $FEHGCD$ is equal to $2\Omega$. Hence we can remove the conducting wire $GH$which is in-between and the equivalent circuit is as shown in the figure given above.
Therefore the Equivalent resistance between $A$ and $B$ is given as ${R_{AB}} = \dfrac{2}{2} = 1\Omega$, which is the answer.

Note:
Wheatstone bridge is a setup that is applicable in finding the unknown resistance. It can be applicable in cases wherein the direct use of Ohm's Law in circuits is not possible. We can also measure minor variations of temperature with the help of thermistors in this Wheatstone bridge. It is also used to find strain and pressure. A photoresistor in this circuit also helps in finding the changes in the intensity of incident light. Because this is a very sensitive device, sometimes the measurements cannot be precise in off-balance conditions. The sensitivity of the circuit will also decrease if the four resistances are dissimilar. It properly works in the small range of ohms to kiloohms.