A) current through its seconary is about four times that of the current through its primary
B) voltage across its secondary is about four times that of the voltage across its primary
C) voltage across its seconary is about two times that of the voltage across its primary
D) voltage across its seconary is about \[\frac{1}{2\sqrt{2}}\] times that of the voltage across its primary.
Correct Answer: A
Solution :
In a transformer the primary and secondary currents are related by \[{{I}_{s}}=\left( \frac{{{N}_{p}}}{{{N}_{s}}} \right){{I}_{p}}\] and the voltages are related by \[{{V}_{s}}=\left( \frac{{{N}_{s}}}{{{N}_{p}}} \right){{V}_{p}}\] where subscripts p and s refer to the primary and secondary of the transformer. Here \[{{V}_{p}}=V,\,\frac{{{N}_{p}}}{{{N}_{s}}}=4\] \[\therefore \,\,\,{{I}_{s}}=4{{I}_{p}}\] \[\therefore \,\,{{I}_{s}}=4{{I}_{p}}\] And \[{{V}_{s}}=\frac{1}{4}V=\frac{V}{4}\]
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