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Q.

An AC source has an internal resistance of 104Ω.  The turn ratio of a transformer so as to match the source to a load of resistance10Ω  is

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a

4.62× 10−2

b

2.03× 10−2

c

3.16× 10−2

d

5.62× 10−2

answer is C.

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Detailed Solution

Power = VS  IS  =  VP IP VS  .   VSZS=VP .  VPZP∴ NsNp=  VSVP=ZSZP  =1010,000  =  3.16  ×  10−2.
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