Question

A galvanometer having 50 divisions provided with a variable shunt S used to measure the current as ammeter when connected in series with a resistance of 90 ohm and a battery of internal resistance 10 ohm, it is observed that when the shunt resistance are 10 and 50 ohm, the deflection are respectively 9 and 30 divisions. what is the resistance of the galvanometer? Further if the full scale deflection of the galvanometer movement is 300mA,find emf of the cell.

Answer 1

hope it helps...................
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Answer 2

Let the emf of the battery is E and resistance of galvanometer is r then,

According to question 

.3×10^−3×9/50=E/90+10r/10+r−−−−(1)

And,

.3×10^−3×3/50=E/90+50r/50+r−−−(2)

Divide first by second

3=90+50r/50+r/90+10r/10+r 

From here you will get value of r

r= −2.43 and −58.81 or

r=2.43,58.81ohm

So, 

For these r there is two possible battery put the value of r in to equation 1 we getE=5mV an 1.8mV respectively for the resistances.