A) 11\[\Omega \]
B) 22\[\Omega \]
C) 99\[\Omega \]
D) 33\[\Omega \]
Correct Answer: A
Solution :
The flowing current in ammeter is 10% of main current. Therefore resistance of shunt connected in parallel. \[{{I}_{1}}{{R}_{1}}={{I}_{2}}{{R}_{2}}\] \[\Rightarrow \]\[\frac{I}{10}\times 99=\frac{9I}{10}\times S\]\[\Rightarrow \]\[S=11\,\Omega \]
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