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The 300 turn primary of a tansformer has resistance 0.82Omega and the resistance of its secondary of 1200 turns is 6.2Omega. Find the voltage across the primary if the poweroutput from the secondary at 1600V is 32 k W. Calculate the power losses in both coils when the transformer effciency is 80%. |
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Answer» Solution :GIVEN : EFFICIENCY `eta=80%` `N_(p)=300,R_(p)=0.82Omega,R_(s)=6.2Omega,N_(s)=1200V,` Output power `=32kW,V_(s)=1600V.` To find : Power loss in the PRIMARY `=I_(p)^(2)R_(p)` Power loss in the sencondary `=I_(s)^(2)R_(s)` `V_(p)=?` Solution : `eta=("Out put power")/(Inpur power")=(V_(s)I_(s))/(V_(p)I_(p))` `(80)/(100)=(32xx10^(-3))/("Inpur Power")` Input Power `=(32xx10^(-3)xx10^(-3))/(80)=40kW.` `(N_(s))/(N_(s))=(V_(s))/(V_(s))` `V_(p)=(V_(s)xxN_(p))/(N_(s))` `V_(p)=(1600xx300)/(1200)=400V` `V_(p)=400V.` Input power `=V_(p)I_(p)` `40xx10^(-3)=400xxI_(p)` `I_(p)=(40000)/(400)=100A` Output power `=V_(s)I_(s)` `32xx10^(3)=1600xxI_(s)` `I_(s)=(32000)/(1600)=20A` Power loss in the primary `=I_(p)^(2)xxR_(p)` `=(100)^(2)xx0.82=8200` = 8.2 k W. Power loss in the secondary `=I_(s)^(2)xxR_(s)` `=(20)^(2)xx6.2=2480` `=2.48kW.` |
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