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Calculate the internal resistance of the given cell using the following data. Balancing length for in open circuit = l _(1) = 0.60m |
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Answer» Solution :`r = R [ (l _(1))/(l _(2)) -1 ]` Trial `(i) r = R [ (l _(1))/( l _(2)) -1 ] =3 (0.6)/(0.4) -1] =1.5 Omega` Trial (ii) `r = R [ ((l _(1))/( l _(2))-1] =4 [ (0.6)/(0.44)-1] =1.45 Omega` Trial (III) `r = R [ (l _(1))/( l _(2)) -1 ] =5 [ (0.6)/(0.46) -1] = 1.52 Omega` Result : Internal resistance of the given cell LIES between `1.45 and 1.52 Omega` |
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