When a current $I$ is passed through a wire of constant resistance, it produces a potential difference $V$ across its ends. The graph drawn between $\log\, I$ and $\log\, V$ will be
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According to ohm's law $V = iR$
$ \Rightarrow $ ${\log _e}V = {\log _e}i + {\log _e}R$ $ \Rightarrow $ ${\log _e}i = {\log _e}V - {\log _e}R$
The graph between ${\log _e}I$ and ${\log _e}V$ will be a straight line which cut ${\log _e}V$ axis and it's gradient will be positive.
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