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whn we convert galvanometer to ammeter we add a shunt to it. d shunt lowers the resistance of the circuit. again as d shunt resistance is very low as compared to that of d galvanometer, only a fraction of current passes thru d galvanometer.
so, my doubt is that- since only a little amount of current is passed thru galvanometer, then how do we obtain the correct reading of current??? (consider the case of moving coil galvanometer).
plz explain using formulae......
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