A) 1000 : 1
B) 100 : 1
C) 1 : 32
D) 32 : 1
Correct Answer: A
Solution :
From Faraday's law, the induced emf across primary and secondary is \[{{e}_{p}}=-{{N}_{p}}\frac{\Delta \text{o}|}{\Delta t}\] \[{{e}_{s}}=-{{N}_{s}}\frac{\Delta \text{o}|}{\Delta t}\] Also, \[e=iR\] \[\therefore \] \[\frac{{{R}_{p}}}{{{R}_{s}}}=\frac{{{N}_{p}}}{{{N}_{s}}}\] Given, \[{{R}_{s}}=8000\,\Omega ,{{R}_{p}}=8\,\Omega \] \[\therefore \] \[\frac{{{N}_{s}}}{{{N}_{p}}}=\frac{{{R}_{s}}}{{{R}_{p}}}=\frac{8000}{8}=\frac{1000}{1}\]
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