Current Electricity Charging Discharging Of Capacitors Question 380
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
[MP PET 1999]
Options:
A) 1 A towards right
B) 0.66 A towards right
C) 0.66 A towards left
D) Zero
Show Answer
Answer:
Correct Answer: B
Solution:
Net current $ i={i _{+}}+{i _{-}}=\frac{({n _{+}}),({q _{+}})}{t}+\frac{({n _{-}}),({q _{-}})}{t} $
Therefore $ i=\frac{({n _{+}})}{t}\times e+\frac{({n _{-}})}{t}\times e $
$ =2.9\times 10^{18}\times 1.6\times {{10}^{-19}}+1.2\times 10^{18}\times 1.6\times {{10}^{-19}} $
Therefore $ i=,0.66,A $
BETA
