A) \[\text{0}\text{.16A,3}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{A}\]
B) \[\text{3}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{A,}\,\text{0}\text{.16A}\]
C) \[\text{0}\text{.16A, }\!\!\times\!\!\text{ }\,\text{0}\text{.16A}\]
D) \[\text{3}\text{.2 }\!\!\times\!\!\text{ 1}\,\text{0}{{\text{.}}^{\text{-3}}}\text{A,}\,\text{3}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{A}\]
Correct Answer: B
Solution :
We have, \[\frac{{{N}_{S}}}{{{N}_{P}}}=\frac{{{i}_{P}}}{{{I}_{S}}}\] \[\frac{10}{500}=\frac{{{i}_{S}}}{{{i}_{S}}}\] \[\Rightarrow \] \[\frac{{{i}_{P}}}{{{i}_{S}}}=\frac{1}{50}\] \[\Rightarrow \] \[{{i}_{s}}=50\,{{i}_{p}}\] This condition is satisfied only when current in primary \[3.2\times {{10}^{-3}}\] A and in secondary 0.16 A.
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