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The current measuring capacity of a galvanometer of resistance `50 Omega` is to be increased from x mA to `101 xxmA`.The shunt resistance isA. `0.5Omega`B. `0.25Omega`C. `1.5Omega`D. `0.35Omega` |
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Answer» Correct Answer - A `S=(G)/((n-1))=(50)/((101-1))=(50)/(100)=0.5Omega` . |
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