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