Question

A single phase Transformer with a ratio of 440/110vtakes a no load current of 5 A at 0.2.pf laggingit the secondary supplies a current of 120A at aPower factor of 0.8 lagging, estimate the currenttaken by the primary


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Answer 1

Answer:

Transformation ratio, K=\frac{E_{2}}{E_{1}}=\frac{110}{440}=0.25K=E1E2=440110=0.25

 

Let the primary counter balancing current be {I_{1}}’I1’.

Then {I_{1}}’=K\, I_{2}=0.25*120=30\: AI1’=KI2=0.25∗120=30A

 

Now cos\: \phi _{o}=0.2;\: \phi _{0}=cos^{-1}\, 0.2=78.46^{\circ}cosϕo=0.2;ϕ0=cos−10.2=78.46∘

 

cos\: \phi _{2}=0.8;\: \phi _{2}=cos^{-1}\, 0.8=36.87^{\circ}cosϕ2=0.8;ϕ2=cos−10.8=36.87∘

 

\theta =\phi _{0}-\phi _{2}=78.46^{\circ}-36.87^{\circ}=41.59^{\circ}θ=ϕ0−ϕ2=78.46∘−36.87∘=41.59∘

 

I_{1}=\sqrt{(I_{0})^{2}+({I_{1}}’)^{2}+2\, I_{0}\, {I_{1}}’\, cos\theta }

=(5)2+(30)2+2∗5∗30∗cos41.59”=33.9A

 

Primary p.f., cos\: \phi _{1}=\frac{{I_{1}’}\, cos\, \phi _{2}+I_{0}\, cos\, \phi _{0}}{I_{1}}cosϕ1=I1I1’cosϕ2+I0cosϕ0

 

=\frac{30*0.8+5*0.2}{33.9}=0.7375\, \, lag\, \,=33.930∗0.8+5∗0.2=0.7375lag

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