Question

A single phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3A at a power factor of 0.2 lag and the secondary current is 280A at a power factor of 0.8 lag. Neglect R₂ and X₂ calculate (1) Magnetizing component and loss component of no load current. (2) primary current (3) input power factor .Draw phasor diagram showing all the currents.

Answer 1

`I_P/I_S=N_S/N_P`

Therefore, `I_P=N_S/N_PxxI_S=200/1000xx280=56A`

cos𝜑₂=0.8 cos𝜑₀=0.2 sin𝜑₂=0.6 sin𝜑₀=0.98
solve for horizontal and vertical components
𝐼₁𝑐𝑜𝑠𝜑₁=𝐼₂𝑐𝑜𝑠𝜑₂+𝐼₀𝑐𝑜𝑠𝜑₀ =(56×0.8)+(3×0.2)=45.4𝐴
𝐼₁𝑠𝑖𝑛𝜑₁=𝐼₂𝑠𝑖𝑛𝜑₂+𝐼₀𝑠𝑖𝑛𝜑₀ =(56×0.6)+(3×0.98) = 36.54A
𝐼₁=`sqrt(45.4^2+36.54^2)`=58.3𝐴
𝑰_𝝁=𝑰_𝟎𝒔𝒊𝒏𝝋_𝟎=(𝟑×𝟎.𝟔)=𝟏.𝟖A

tan𝜑₁ `(36.54)/(45.4)`= 0.805

𝛗_𝟏=𝟑𝟖°
Power factor cos𝝋_𝟏=𝒄𝒐𝒔𝟑𝟖°=𝟎.𝟕𝟖 lagging.