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Two identical cells either in series or in parallel combination , gives the same current of 0.5 Athrough external resistance of 4Omega. Find emf and internal resistance of each cell. |
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Answer» Solution :Let each cell be of emf E and internal resistance r, `R=4Omega`, I=0.5 A When the two cells are connected in series : Total emf =E+E=2E , Total resistance = 4+r+r =4+2r `therefore` Circuit current `I=E/(R+r)` `therefore I_1=(2E)/(4+2r) to ` (1) When the two cells are connected in PARALLEL : Total emf =E , Total resistance =`4+r/2` Circuit current `I_2=E/(4+r/2) =(2E)/(8+r) to` (2) As `I_1=I_2` 8+r=4+2r 4=r `r=4Omega` eqn. (1) `RARR` 0.5 =`(2E)/(4+2(4)) rArr 2+4=2E rArr 6/2 =E rArr E=3V` |
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