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The resistance of an ammeter is `13 Omega` and its scale is graduated for a current upto `100 A`. After an additional shunt has been connected to this ammeter it becomes possible to measure currents upto `750 A` by this meter. The value of shunt resistance isA. `20 Omega`B. `2Omega`C. `0.2Omega`D. `2kOmega` |
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Answer» Correct Answer - B Let `i_(a)` is the current flowing through ammeter and i is the total current. So, a current `i-i_(a)` will flow through shunt resistance. Potential difference across ammeter and shunt resistance is same. i.e., `i_(a)xx R=(i-i_(a))xxS` or `" " S=(i_(a)R)/(i-i_(a))` ....(i) Give, `i_(a)=100 A,i=750 A, R=13 Omega` Hence, `S=(100xx13)/(750-100)=2 Omega` |
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