A) 10 ohm
B) 9.9 ohm
C) 9 ohm
D) 11 ohm
Correct Answer: D
Solution :
Using the relation for shunt resistance \[S=\frac{{{i}_{g}}}{i-{{i}_{g}}}\times G\] ...(1) Where S = resistance of shunt G = resistance of galvanometer i = total current \[{{i}_{g}}\]= current for full scale deflection in (galvanometer) Here \[{{i}_{g}}=\frac{10}{100}i=\frac{i}{10}G=99\Omega ,\,\,\,S=?\] From eqn. (1) \[S=\frac{i\times 99}{10\left( i-\frac{i}{10} \right)}=\frac{99}{10\times \frac{9}{10}}=11\Omega \]
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