The wire mesh shown is made of a uniform wire. Resistance between any two points is $$R$$. When wires $$OH$$ and $$OD$$ are removed some current flows between the terminals when a potential difference $$V$$ is applied between them. If now the wires $$OB$$ and $$OF$$ are also removed, find the change in potential difference for same current to flow between the terminals.
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Each segment between adjacent points has resistance $$R$$ and the terminals are at $$A$$ and $$G$$. With $$OH$$ and $$OD$$ removed, the path $$A\text{-}H\text{-}G$$ is $$2R$$, and the lower path is $$AB$$, then $$B\text{-}O\text{-}F$$ ($$2R$$) in parallel with $$B\text{-}C\text{-}D\text{-}E\text{-}F$$ ($$4R$$), then $$FG$$, giving $$R + \frac{4R}{3} + R = \frac{10R}{3}$$; in parallel with $$2R$$ this gives $$\frac{5R}{4}$$, so the current is $$I = \frac{4V}{5R}$$. With $$OB$$ and $$OF$$ also removed, the lower path is $$6R$$ and the total resistance is $$\frac{2R \times 6R}{8R} = \frac{3R}{2}$$. The same current then needs $$\frac{4V}{5R} \times \frac{3R}{2} = \frac{6V}{5}$$, so the potential difference must increase by $$\frac{V}{5}$$.