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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`.A. 0.234 AB. 0.632 AC. 0.857 AD. None of these |
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Answer» Correct Answer - C 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` |
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