Seven resistances are connected as shown in the figure. The equivalent resistance between $A$ and $B$ is ............... $\Omega$
Medium
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(b) The given circuit can be simplified as follows
So, ${R_{AB}} = \frac{{8 \times 8}}{{8 + 8}} = \frac{{64}}{{16}} = 4\Omega $
So, ${R_{AB}} = \frac{{8 \times 8}}{{8 + 8}} = \frac{{64}}{{16}} = 4\Omega $
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