First slide

electric current

Question

In a neon discharge tube $2.9 \times 10^{18} N e^{+}$ ions move to the right each second while $1.2 \times 10^{18}$ electrons move to the left per second. Electron charge is $1.6 \times 10^{- 19} C .$ The current in the discharge tube

Easy

A

1 A towards right
B

0.66 A towards right
C

0.66 A towards left
D

Zero

Solution

$N e t c u r r e n t i = i_{+} + i_{-} = \frac{(n_{+}) (q_{+})}{t} + \frac{(n_{-}) (q_{-})}{t}$
$\begin{array}{l} N e t c u r r e n t i = i_{+} + i_{-} = \frac{(n_{+}) (q_{+})}{t} + \frac{(n_{-}) (q_{-})}{t} \\ \Rightarrow i = \frac{(n_{+})}{t} \times e + \frac{(n_{-})}{t} \\ = 2.9 \times 10^{18} \times 1.6 \times 10^{- 19} + 1.2 \times 10^{18} \times 1.6 \times 10^{- 19} \\ \Rightarrow i = 0.66 A . \end{array}$

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