A) 0.08 A
B) 0.4 A
C) 5 A
D) 10 A
Correct Answer: D
Solution :
\[\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{I}_{p}}}{{{I}_{s}}}\Rightarrow \frac{20}{100}=\frac{2}{{{I}_{s}}}\] \[=\frac{1}{5}=\frac{2}{{{I}_{s}}}\] \[\therefore \]The current in the secondary coil \[{{I}_{s}}=5\times 2=10A\]You need to login to perform this action.
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