In an electric circuit $1 \times 10^{18}$ hydrogen ions are travelling per second in the right direction while double number of electrons are travelling per second in left direction, the total current through the path is _________.(Charge on one electron $= 1.6 \times 10^{- 19}$ C).

4.8

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1

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0.48

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0.1

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Solution

The correct option is C $0.48$ A
Number of hydrogen ions moving per sec towards right $n_{1} = 1 \times 10^{18}$
Electric current due to hydrogen ions $I_{1} = n_{1} e$
$⟹$ $I_{1} = 1 \times 10^{18} \times 1.6 \times 10^{- 19} = 0.16$ A (towards right)
Number of electrons moving per sec towards left $n_{2} = 2 n_{1} = 2 \times 10^{18}$
Electric current due to electrons $I_{2} = n_{2} e$
$⟹$ $I_{2} = 2 \times 10^{18} \times 1.6 \times 10^{- 19} = 0.32$ A (towards right)
Thus total current in the circuit $I = I_{1} + I_{2}$
$⟹$ $I = 0.16 + 0.32 = 0.48$ A

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In an electric circuit 1×1018 hydrogen ions are travelling per second in the right direction while double number of electrons are travelling per second in left direction, the total current through the path is _________.(Charge on one electron=1.6×10−19C).

In an electric circuit $1 \times 10^{18}$ hydrogen ions are travelling per second in the right direction while double number of electrons are travelling per second in left direction, the total current through the path is _________.(Charge on one electron $= 1.6 \times 10^{- 19}$ C).