If 8 cells having an emf of 1.5 V are connected in series across a load having resistance of `10 Omega`. What is the current drawn by the load ? Assume internal resistance of all be `0.5 Omega`.

Options

A. 0.234 A
B. 0.632 A
C. 0.857 A
D. None of these

Answer

C

Explanation

Given, emf of each cell, E = 1.5 V
Internal resistance of each cell, `r = 0.5 Omega`
Number of cells, n = 8
Load resistance, `R = 10 Omega`
Since, cells are connected in series
`therefore " "` Total emf, `E_(T)=nE`
`= 8xx1.5=12 V`
Total resistance, `R_(T)=nr+R`
`= (8xx0.5)+10`
`= 14 Omega`
Total circuit current, `I=(E_(T))/(R_(T))`
`= (12)/(14)=0.857 A`