A) 0.1 \[\Omega \] in parallel
B) 0.1 \[\Omega \] in series
C) 0.2 \[\Omega \] in parallel
D) 0.2 \[\Omega \] is series
Correct Answer: A
Solution :
Here: resistance of galvanometer\[G=100\,\Omega \] Maximum current across galvanometer \[{{I}_{g}}=0.01A\] Current range in ammeter = 10 A Since, the shunt resistance should be connected in parallel to convert the galvanometer into an ammeter. \[S=\left( \frac{{{I}_{g}}}{I-{{I}_{g}}} \right)G=\left[ \frac{0.01}{(10-0.01)} \right]\times 100\] \[=0.1\,\Omega \]
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