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Two cells of emf 2 epsi and epsi and internal resistance 2r and r respectively, are connected in parallel.Obtain the expressions for the equivalent emf and the internal resistance of the combination. |
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Answer» Solution :Here `epsi_1 = 2 epsi , espi_2 =epsi, r_1 = 2R` and `r_2=r ` and the two cells have been connected in parallel. If equivalent emf and INTERNAL RESISTANCE of the COMBINATION be `epsi_(eq)` and `r_(eq)`respectively, then `(epsi_(eq))/(r_(eq)) = epsi_1/r_1+ epsi_2/r_2 " and " (1)/(r_(eq)) = (1)/(r_1) + (1)/(r_2)` ` therefore r_(eq) = (r_1 r_2)/(r_1 + r_2) = ((2r)(r ) )/(2r + r) = 2/3 r` and `(epsi_(eq))/( (2/3 r) )= (2 epsi)/(2 r) + epsi/r = (2epsi)/(r ) rArr epsi_(eq) = 2/3 r xx (2 epsi)/(r ) = 4/3 epsi` |
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