1.

What is the value of shunt which passes 10% of the main current through a galvanometer of 99 ohm?

Answer»

SOLUTION :As shunt is a small resistance S in parallel with a galvanometer (of resistance G) as shown in FIG.
`(I-I_(g))S=I_(G)G`
i.e. `S=(I_(G)G)/((I-I_(G)))`
And as here, `G=99Omega` and
`I_(G)=(10/100)I=0.1IS=(0.1Ixx99)/((I-0.1I))=(0.1)/(0.9)xx99=11Omega`


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