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A 10 kW transformer has 20 turns in primary and 100 turns in secondary circuit. A.C. voltage `E_(1) = 600 sin 314 t` is applied to the primary. Find max. value of flux and max. value of secondary voltage. |
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Answer» i) Flus linked with each turn of primary, `phi=Bacosomegat=phi_(0)cosomegat` Here, `phi_(0)` BA=maximum value of flux linked with each turn `therefore V_(1)=-N_(1)(dphi)/(dt) = -N_(1)d/(dt)(phi_(0)cosomegat) = omegaN_(1)phi_(0)sinomegat` Peak value of `V_(1) = V_(0) = omegaN_(1)phi_(0)` or `phi_(0) = V_(0)/(omegaN_(1))` Given, `V_(1) = 600 sin 314t=V_(0)Sinomegat` `therefore V_(0)=600V, omega=314 rad s^(-1)` Here, `phi^(0) = 600/(314xx20)=0.0955Wb` ii) `V_(2)^(0)/V_(1)^(0) = N_(2)/N_(1)` `therefore` Maximum value of secondary voltage is, `V_(2)^(2) = N_(2)/N_(1)V_(1)^(0)= 100/20 xx 600 = 3000V` |
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