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The number of turns in the primary and secondary coils of an ideal transformer are 2000 and 50 respectively. The primary coil is conneted to a main supply of 120 V and socondary to a night bulb of `0.6 Omega`. Calculate (i) Voltage acorss the secondary. (ii) Current in the bulb, (iii) Current in primarya coil, (iv) Power in primary and secondary coils, (iv) Power in primary and secondary coils. |
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Answer» Here, `n_(p) = 2000, n_(s) = 50`, `E_(p) = 120 V , R_(s) = 0.6 Omega` `E_(s) = ?, I_(s) = ?, I_(p) = ?, P_(p) = ?, P_(s) = ?` (i) As `(E_(s))/(E_(p)) = (n_(s))/(n_(p))` `:. E_(s) = E_(p) . (n_(s))/(n_(p)) = 120 xx (50)/(2000) = 3 V` (ii) As `I_(s) = (E_(s))/(R ) :. I_(s) = (3)/(0.6) = 5 A` (iii) As `(I_(p))/(I_(s)) = (E_(s))/(E_(p))` `:. I_(p) = (E_(s))/( E_(p)) xx I_(s) = (3)/(120) xx 5 = 0.125 A` (iv) Power in primary, `P_(p) = E_(p) xx I_(p) = 120 xx 0.125 = 15 W` Power in secondary, `P_(s) = E_(s) xx I_(s) = 3 xx 5 = 15 W` |
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