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A rangeof galvanometer is V when 50 Omega resistance is connected is connected in series its rangegets doubled when 500 Omegaresistance is connectedin series galvanometerresistance is |
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Answer» `100 omega` If I is maximum current flowingthroughgalvanometer then `V= (R+G)I=(50+G)I` where V is the rangeof galvanometerwhen R= 50 `omega`is connectedin serieswith itand g isgalvanometer resistance when 500 `omega`resistnce is conncected in seriesthe rangebecomesdouble `2V=(500 +G)I` or `500 +G =100 +2Grarr G= 400 omega` |
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