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Using the equation of power for an ideal transformer, prove (I_(p))/( I_(s)) = ( V_(s))/( V_(p)) = ( N_(s))/( N_(p)) |
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Answer» Solution :If the transformer is an ideal one, its efficiency is 100% so no energy losses. The POWER input is equal to the power output and since. p = IV `:. `In put power = Output power `I_(p) V_(p) = I_(s) V_(s) `…(1) Some energy is always lose, since a well designed transformer may have an efficiency of more than 95%. For an ideal transformer, `(V_(s))/(V_(p)) = ( N_(s))/( N_(p)) `...(2) Where `V_(s)` and `V_(p)` voltage across secondary and primary coil respectively and `N_(s)` and `N_(p)` are the number of turns in secondary coil and primary coil respectively. From equation (1) and (2), `(I_(p))/( I_(s)) = (V_(s))/(V_(p)) = ( N_(s))/( N_(p))`...(3) Hence, I and V both OSCILLATE with the same frequency as the ac source equation (3) also gives the ratio of the amplitude or rms value of corresponding quantities. |
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