In an electrolyte 3-2 - 1018 bivalent positive ions drift to the right per second while 3-6 - 1018 monovalent negative ions drift to the left per second- Then the current is

The correct option is A 1.6 amp to the rightNet current i_net=i_(+)+i_( ) =n_(+)q_(+)/t+n_( )q_( )/t =n_(+)/t× 2e+n_( )/t× e =3.2× 10^18× 2× 1.6 × 10^ 19+3.6 ...

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Standard XII

Physics

The Flow of Electrons as Electric Current

In an electro...

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

1.6 amp to the left

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1.6 amp to the right

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0.45 amp to the right

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0.45 amp to the left

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Solution

The correct option is A. 1.6 amp to the right
Net current $i_{n e t} = i_{(+)} + i_{(-)}$
= $\frac{n_{(+)} q_{(+)}}{t} + \frac{n_{(-)} q_{(-)}}{t}$
= $\frac{n_{(+)}}{t} \times 2 e + \frac{n_{(-)}}{t} \times e$

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

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In an electrolyte 3.2× 1018 bivalent positive ions drift to the right per second while 3.6× 1018 monovalent negative ions drift to the left per second. Then the current is

The correct option is A. 1.6 amp to the rightNet current inet=i(+)+i(−) =n(+)q(+)t+n(−)q(−)t =n(+)t× 2e+n(−)t× e =3.2× 1018× 2× 1.6 × 10−19+3.6× 1018× 1.6 × 10−19 = 1.6 A (towards right).

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