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A cell of emfepsiand internal resistance 'r' is connected across a variable load resistor 'R'. Draw the plots of the terminal voltage 'V' versus (i) R and (ii) the current I. It is found that when R = 4 Omega , the current is 1 A and when R is increased to 9 Omega , the current reduces to 0.5 A. Find the values of the emf epsiand internal resistance r. |
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Answer» Solution :We know that current`I = (epsi)/(R + r)` When `R = 4 Omega , I = 1A , ` HENCE`1 = (epsi)/( 4+ r)`.....(i) Again if `R. = 9Omega` then `I. = 0.5A` , hence ` 0.5 = (epsi)/(9 + r)`..(II) From (i) and (ii) , we get , r= 1 `Omega` and ` epsi` = 5V |
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