1.

Calculate the value of the shunt which when connected across a galvanometer of resistance 38 Ω will allow 1/20th of the current to pass through the galvanometer

Answer»

Data : G = 38 Ω, 

\(\cfrac{I_g}I\) = \(\cfrac1{20}\)

Ig\(\left(\cfrac{S}{S+G}\right)I\)

∴ \(\cfrac{I_g}I\) = \(\cfrac{S}{S+G}\)

∴ \(\cfrac1{20}\) = \(\cfrac{S}{S+38}\)

∴ S + 38 = 20 S 

∴ 19 S = 38 

∴ S = 2 Ω

This is the required value of the shunt.



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