A galvanometer having a coil of resistance $30 \Omega$ need 20 mA of current for full-scale deflecti

Question

A galvanometer having a coil of resistance $30 \Omega$ need 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\frac{30}{X} \Omega$, where $X$ is\

Professor PC Thomas Classes · Answer

(d) : So, potential difference across galvanometer wire and shunt will be same.\ $	herefore \quad I_{g} R_{g}=I_{s} R_{s}$\ $I_{g} R_{g}=\left(I-I_{g}ight) R_{s}$\ $\Rightarrow \quad 0.02 	imes 30=2.98 	imes R_{s}$\ $\Rightarrow \quad R_{s}=\frac{0.02 	imes 30}{2.98}=\frac{30}{149} \Omega$\ By comparing with question, $X$ is 149 .\