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

Answer 1

Answer: (B)

Net current $i_{\text {net }}=i_{(+)}+i_{(-)}$

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