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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 |
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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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