InterviewSolution
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InterviewSolution
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What shunt resistance is required to make the `1.00 mA`, `20 Omega` Galvanometer into an ammeter with a range of 0 to `50.0A`? |
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Answer» We have `I_(2) = 1.00mA=1.00xx 10^(-3) A` and `R_(2) = 2.0.0 Omega`, and the ammetre should be able to handle maximum current , `I = 50.0 xx 10^(-3) A`. Solving equation `I_(g) r_(g) = (I - I_(g)) R_(S)` for `R_(S)`, we get `R_(S) = (I_(g)R_(g))/(I - I_(g)) = ((1.00 xx 10^(-3) A)(20.0 Omega))/(50.0 xx 10^(-3) A- 1.00 xx 10^(-3) A) = 0.408 Omega` |
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