A) 0.5 \[\Omega \] in series
B) 0.5 \[\Omega \] in parallel
C) 50 \[\Omega \] in series
D) 50 \[\Omega \] in parallel
Correct Answer: B
Solution :
Here: Resistance of galvanometer \[G=50\Omega \] Current through galvanometer \[{{i}_{g}}=100\mu A\] \[=100\times {{10}^{-6}}={{10}^{-4}}A\] \[i=10\,mA=10\times {{10}^{-3}}=100\times {{10}^{-4}}A\] Shunt required is \[S=\left( \frac{{{i}_{g}}}{i-{{i}_{g}}} \right)G\] \[S=\frac{{{10}^{-4}}}{(100\times {{10}^{-4}}-{{10}^{4}})}\times 50=\frac{{{10}^{-4}}\times 50}{99\times {{10}^{-4}}}\] \[\approx 0.5\,\Omega \] The shunt resistance is connected in parallel to convert galvanometer into ammeter.
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