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 is \[{{I}_{g}}=1\,mA=1\times {{10}^{-3}}A\] Now, for shunt there 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 \[R=\frac{{{I}_{g}}G}{I-{{I}_{g}}}\] ?(i) (where R is shunt resistance \[R=4\Omega \]) Now, putting the values in Eq. (i) \[4=\frac{1\times {{10}^{-3}}\times 20}{I-1\times {{10}^{-3}}}\] \[4I-4\times {{10}^{-3}}=20\times {{10}^{-3}}\] \[4I=20\times {{10}^{-3}}+4\times {{10}^{-3}}=24\times {{10}^{-3}}\] \[I=\frac{24\times {{10}^{-3}}}{4}=6\times {{10}^{-3}}A\] \[=6\,mA\]
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