A bridge circuit is shown in figure. The equivalent resistance between $$A$$ and $$B$$ will be

Solution :
The bridge circuit can be shown as

The balanced condition of bridge circuit is given by
$$\frac{P}{Q} = \frac{3}{4},\frac{R}{S} = \frac{6}{8} = \frac{3}{4} \Rightarrow \frac{P}{Q} = \frac{R}{S}$$
Thus, it is balanced Wheatstone bridge, so potential at $$F$$ is equal to potential at $$H.$$ Therefore, no current will flow through $$7\,\Omega $$  resistance. So, circuit can be redrawn as

$$P$$ and $$Q$$ are in series, so their equivalent resistance
$$ = 3 + 4 = 7\,\Omega $$
$$R$$ and $$S$$ are also in series, so their equivalent resistance $$ = 6 + 8 = 14\,\Omega $$
Now, $$7\,\Omega $$  and $$14\,\Omega $$  resistances are in parallel, so
$${R_{AB}} = \frac{{7 \times 14}}{{7 + 14}} = \frac{{7 \times 14}}{{21}} = \frac{{14}}{3}\Omega $$
NOTE
Normally, in Wheatstone bridge in middle arm galvanometer must be connected. In Wheatstone bridge, cell and galvanometer arms are interchangeable. In both the cases, condition of balanced bridge is
$$\frac{P}{Q} = \frac{R}{S}$$