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A galvanometer has a resistance of 98Omega. 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=98Omega,(i_(g))/ixx100=2%` `s=G/((i/(i_(g))-1)),thereforei/(i_(g))=100/2=50thereforeS=98/((50-1))=2OMEGA` |
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