A) 10
B) 8
C) 6
D) 4
Correct Answer: C
Solution :
Given, Meter has a resistance \[G=20\,\Omega \] Current through meter for full deflections \[{{l}_{g}}=1mA=1\times {{10}^{-3}}A\] Now, for shunt three resistors each have resistance \[12\,\Omega \] are connected in parallel. Hence, \[\frac{1}{{{R}_{p}}}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}\Rightarrow {{R}_{p}}=4\,\Omega \] Now, from the formula \[R=\frac{{{l}_{g}}G}{l-{{l}_{g}}}\] ? (i) (where, R is shunt resistance \[R=4\,\Omega \]) Now, putting the values in Eq. (i) \[4=\frac{1\times {{10}^{-3}}\times 20}{l-1\times {{10}^{-3}}}\] \[4l-4\times {{10}^{-3}}=20\times {{10}^{-3}}\] \[4l=20\times {{10}^{-3}}+4\times {{10}^{-3}}=24\times {{10}^{-3}}\] \[l=\frac{24\times {{10}^{-3}}}{4}=6\times {{10}^{-3}}A=6mA\]
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