A) 1A
B) 4.5 A
C) 2 A
D) 1.5A
Correct Answer: C
Solution :
Key Idea: For an ideal transformer output power = input power, If\[{{i}_{s}}\]and\[{{i}_{p}}\]be the currents in the primary and secondary at any instant and the energy losses be zero, then power in secondary = power in primary \[{{V}_{s}}\times {{i}_{s}}={{V}_{p}}\times {{i}_{p}}\] \[\Rightarrow \]\[\frac{{{i}_{p}}}{{{i}_{s}}}=\frac{{{V}_{s}}}{{{V}_{p}}}=\frac{{{N}_{s}}}{{{N}_{p}}}=\]transformer ratio. Given, \[\frac{{{N}_{p}}}{{{N}_{s}}}=\frac{2}{3},{{i}_{p}}=3A\] \[\Rightarrow \] \[{{i}_{s}}=\frac{{{N}_{p}}}{{{N}_{s}}},{{i}_{p}}=\frac{2}{3}\times 3=2A\] Note: When voltage is stepped-up the current is correspondingly reduced in the same ratio and vice-versa.
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