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Two cells of emf epsi_1 and epsi_2having internal resistances r_1 and r_2respectively are connected inparallel as shown in Fig. Deduce the expressions for the equivalent emf and equivalent internal resistance of a cell which can replace the combination between the points B_1 and B_2 . |
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Answer» Solution :Consider a parallel combination of two cells. Let `epsi_1` and `epsi_2`be theemfs of two cells and `r_1` and `r_2 ` their respective internal resistances. `I_1` and `I_2`are the currents leaving the positive ELECTRODES of two cells and at junction `B_1`, these currents are added up so that net current `I = I_1 + I_2`.....(i) Let `V(B_1)` and `V(B_2)`be the potentials at `B_1` and `B_2` , RESPECTIVELY, then considering first cell, we have ` V = V(B_1) = V(B_2) = epsi_1 - I_1 r_1 rArr I_1 = (epsi_1 - V)/(r_1)`...(ii) and considering second cell, we have ` V = V(B_1) - V(B_2) = epsi_2 - I_2 r_2 rArr I_2 = (epsi_2 - V)/(r_2)` ` therefore I = I_1+ I_2 = (epsi_1 - V)/(r_1) + (epsi_2- V)/(r_2) = ( epsi_1/r_1 + epsi_2/r_2) - V ((1)/(r_1) - (1)/(r_2))` `rArr V = [ ( espi_1/r_1+ espi_2/r_2) - I ] // ( (1)/(r_1) + (1)/(r_2) ) = ((epsi_1 r_2 + epsi_2 r_1)/(r_1 + r_2))- I ( (r_1 r_2)/(r_1 + r_2))`...(iv) If instead of two cells we join a single cell having equivalent emf `epsi_(eq)`and equivalent internal resistance `r_(eq)` , then we have `V = epsi_(eq) - Ir_(eq)`....(v) Thus, comparing (iv) and (v), we get `epsi_(eq) = (( epsi_1 r_2 + epsi_2 r_1)/(r_1 + r_2)) ` and ` r_(eq) = (r_1 r_2)/(r_1 + r_2)` Alternately , we have can WRITE `(1)/(r_(eq)) = (1)/(r_1) + (1)/(r_2) ` and `( epsi_(eq))/(r_(eq)) = epsi_1/r_1 + epsi_2/r_2` |
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