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A galvanometer has a resistance of 98 Omega. If 2 % of the main current is to be passed through the meter , what should be the value of the shunt ? |
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Answer» SOLUTION :`G = 98 OMEGA , (i_g)/(I) = 2 %` `s= (G)/(((i)/(i_g)-1)) , therefore i/(i_g) = 100/2 = 50` `thereforeS = (98)/((50-1)) = 2 Omega` |
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