Current Electricity Charging Discharging Of Capacitors Question 49
Question: If in a voltaic cell 5 gm of zinc is consumed, then we get how many ampere hours ? (Given that E.C.E. of Zn is $ 3.387\times {{10}^{-7}} $ kg/coulomb)
Options:
A) 2.05
B) 8.2
C) 4.1
D) $ 5\times 3.387\times {{10}^{-7}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ m=Zi,t=Zq $ ; $ q=\frac{5\times {{10}^{-3}}}{3.387\times {{10}^{-7}}}amp\text{-}sec $ or $ q=\frac{5\times {{10}^{-3}}}{3.387\times {{10}^{-7}}\times 3600}amp\text{-}hr=4.1 $
BETA
