The amount of current flowing through the channel is found by:

$\overline{){\mathbf{i}}{\mathbf{=}}{\mathbf{(}}{\mathbf{i}}{\mathbf{o}}{\mathbf{n}}{\mathbf{s}}{\mathbf{}}{\mathbf{p}}{\mathbf{e}}{\mathbf{r}}{\mathbf{}}{\mathbf{s}}{\mathbf{e}}{\mathbf{c}}{\mathbf{o}}{\mathbf{n}}{\mathbf{d}}{\mathbf{)}}{\mathbf{\times}}{\mathbf{(}}{\mathbf{c}}{\mathbf{h}}{\mathbf{a}}{\mathbf{r}}{\mathbf{g}}{\mathbf{e}}{\mathbf{}}{\mathbf{o}}{\mathbf{f}}{\mathbf{}}{\mathbf{e}}{\mathbf{a}}{\mathbf{c}}{\mathbf{h}}{\mathbf{}}{\mathbf{i}}{\mathbf{o}}{\mathbf{n}}{\mathbf{)}}}$

Power:

$\overline{){\mathbf{P}}{\mathbf{=}}{\mathbf{i}}{\mathbf{\u2206}}{\mathbf{V}}}$

When an ion channel opens in a cell wall (see Problem 40), monovalent (charge e) ions flow through the channel at a rate of 2.9x10^{7} ions/s. (a) What is the current through the channel? (b) The potential difference across the ion channel is 88 mV. What is the power dissipation in the channel?

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