Question

In an electrolyte $3.2\times {10}^{18}$ bivalent positive ions drift to the right per second while $3.6\times {10}^{18}$ monovalent negative ions drift to the left per second. Then the current is

• A. 1.6 amp to the left
• B. 1.6 amp to the right
• C. 0.45 amp to the right
• D. 0.45 amp to the left

Answer 1

The correct option is B 1.6 amp to the right

Net current $i_{net}=i_{(+)}+i_{(-)}$
=$\frac{n_{(+)}{q}_{(+)}}{t}+\frac{n_{(-)}{q}_{(-)}}{t}$
=$\frac{n_{(+)}}{t}\times 2e+\frac{n_{(-)}}{t}\times e$


$=3.2\times {10}^{18}\times 2\times 1.6\times {10}^{-19}+3.6\times {10}^{18}\times 1.6\times {10}^{-19}$
=1.6 A (towards right)