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STD 12 - 3. current electricity — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3. current electricity4 Marks
MCQ
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
  • ✓
    $1.6\,amp$ to the right
  • C
    $0.45\,amp$ to the right
  • D
    $0.45\,amp$ to the left
✓

Answer

Correct option: B.
$1.6\,amp$ to the right
b
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)

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