In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?

In a galvanometer there is a deflection of 10 divisions per `50 mA`. T

1.	In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?
Answer» Correct Answer - (a) Increasing the turns by 20% (b) Lesser Since the galvanometer has 50 divisions, so current for full scale deflection is `I_g=1/10xx50mA=5mA=5xx10^-3A`, `G=60Omega`, `S=25Omega`. Let I be the maximum current which a galvanometer can read when shunted with resistance S, then `I_g=(IS)/(G+S)` or `I=(I_g(G+S))/(S)` `=((5xx10^-3)(60+25))/(2*5)` `=125xx10^-3A` `=125mA`

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In a galvanometer there is a deflection of 10 divisions per `50 mA`. The internal resistance of the galvanometer is `60Omega`. If a shunt of `2*5Omega` is connected to the galvanometer and there are 50 divisions in all on the scale of galvanometer what maximum current can this galvanometer read?

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