Transformer Phasor Diagrams: Solved Problems

Problem-1

A 2200/200-V transformer draws a no-load primary current of 0.6 A and absorbs 400 watts. Find the magnetizing and iron loss currents.

Solution-1

\[\begin{aligned} \text{Iron-loss current},~I_w & = \dfrac{W_0}{V_1} \\ & = \dfrac{400}{2200} = 0.182 ~ \mathrm{A}\\ \\ I_0^2& =I_w^2+I_{\mu}^2 \\ \Rightarrow I_{\mu}& = \sqrt{\left(0.6^2-0.182^2\right)}=0.572~\mathrm{A} \end{aligned}\]

Problem-2

A 2200/2540 V transformer takes 0.5 A at a power factor of 0.3 on open circuit. Find magnetizing and working components of no load primary current.

Solution-2

\[\begin{aligned} I_{0}&=0.5 \mathrm{~A} \\ \cos \phi_{0}&=0.3 \\ \\ I_{w}&=I_{0} \cos \phi_{0}=0.5 \times 0.3=0.15 \mathrm{~A} \\ I_{\mu}&=\sqrt{0.5^{2}-0.15^{2}}=0.476 \mathrm{~A} \end{aligned}\]

Problem-3

A single-phase transformer has 500 turns on the primary and 40 turns on the secondary winding. The mean length of the magnetic path in the iron core is 150 cm and the joints are equivalent to an air-gap of 0.1 mm. When a voltage of 3000 V is applied to the primary, maximum flux density is 1.2 Wb/m\(^2\). Calculate

the cross-sectional area of the core
no-load secondary voltage
the no-load current drawn by the primary
power factor on no-load.

Given that AT/cm for a flux density of 1.2 Wb/m\(^2\) in iron to be 5, the corresponding iron loss to be 2 watt/kg at 50 Hz and the density of iron as 7.8 gram/cm3.

Solution-3

cross-sectional area of the core \[\begin{aligned} E_1,~3000&=4.44 \times 50 \times 500 \times 1.2 \times \mathrm{A} \\ A&=0.0225 \mathrm{~m}^{2}\\ &=225 \mathrm{~cm}^{2} \end{aligned}\]
no-load secondary voltage \[\begin{aligned} K & = N_2/N_1 = 40/500 = 4/50 \\ E_2 & = KE_1 = (4/50) \times 3000 = 240~\mathrm{V} \end{aligned}\]

the no-load current drawn by the primary \[\begin{aligned} \text{AT per cm} &=5 \\ \text{AT for iron core} &=150 \times 5=750 \\ \text{AT for air-gap} & =H I=\frac{B}{\mu_{0}} \times l\\ &=\frac{1.2}{4 \pi \times 10^{-7}} \times 0.0001=95.5 \\ \text{Total AT for given}~ B_{\max }&=750+95.5=845.5 \\ \end{aligned}\]

\[\begin{aligned} \text{Max. value of }~I_{\mu} &=845.5 / 500=1.691 \mathrm{~A} \\ \text{r.m.s. value of}~ I_{\mu}&=1.691 / \sqrt{2}=1.196 \mathrm{~A}\\ \text{Volume of iron} &= \text{length} \times \text{area}\\ & =150 \times 225=33.750 \mathrm{~cm}^{3} \\ \text{Density} &=7.8 \mathrm{gram} / \mathrm{cm}^{3} \\ \text{ Mass of iron} & =33,750 \times 7.8 / 1000=263.25 \mathrm{~kg}\\ \text{Total iron loss} &=263.25 \times 2=526.5 \mathrm{~W}\\ I_{w} & =526.5 / 3000=0.176 \mathrm{~A} \\ I_{0}&=\sqrt{I_{\mu}^{2}+I_{w}^{2}}=\sqrt{1.196^{2}+0.176^{2}}=0.208 \mathrm{~A} \end{aligned}\]

power factor on no-load. \[\cos\phi_0 = I_w/I_0 = 0.176/1.208=0.1457\]

Problem-4

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3 A at a power factor of 0.2 lagging. Calculate the primary current and power factor when the secondary current is 280 A at a power factor of 0.80 lagging.

Solution-4

\(V_{2}\) is taken as reference. \[\begin{aligned} \cos ^{-1} 0.80 & =36.87^{\circ}\\ I_{2} &=280 \angle-36.87^{\circ} \text { amp } \\ I_{2}^{'} &=(280/5) \angle-36.87^{\circ} \mathrm{amp} \\ \phi &=\cos ^{-1} 0.20=78.5^{\circ} \\ \sin \phi & =0.98 \\ I_{1} &=I_{0}+I_{2}^{'}\\ &=3(0.20-j 0.98)+56(0.80-j 0.60) \\ &=45.4-j 36.54\\ &=58.3 \angle -38.86^{\circ} \end{aligned}\] Thus \(I\) lags behind the supply voltage by an angle of \(38.86^{\circ} .\)

Problem-5

A single-phase transformer with a ratio of 440/110 V takes a no-load current of 5 A at 0.2 power factor lagging. If the secondary supplies a current of 120 A at a power factor of 0.8 lagging, estimate the current taken by the primary.

Solution-5

\[\begin{aligned} \cos \phi_{2} &=0.8 \\ \phi_{2}&=\cos ^{-1}(0.8)=36^{\circ} 54^{\prime} \\ \cos \phi_{0}&=0.2 \\ \phi_{0}&=\cos ^{-1}(0.2)=78^{\circ} 30^{\prime} \\ K&=V_{2} / V_{1}=110 / 440=1 / 4 \\ I_{2}^{\prime} & =K I_{2}=120 \times 1 / 4=30 \mathrm{~A} \\ I_{0} & =5 \mathrm{~A} \end{aligned}\]

\[\begin{aligned} \text{Angle between}~ I_{0}~ \text{and}~ I_{2}{ }^{\prime} \\ &=78^{\circ} 30^{\prime}-36^{\circ} 54^{\prime}=41^{\circ} 36^{\prime}\\ I_{1} &=\sqrt{\left(5^{2}+30^{2}+2 \times 5 \times 30 \times \cos 41^{\circ} 36^{\prime}\right)} \\ &=34.45 \mathrm{~A} \end{aligned}\]

Problem-6

A transformer has a primary winding of 800 turns and a secondary winding of 200 turns. When the load current on the secondary is 80 A at 0.8 power factor lagging, the primary current is 25 A at 0.707 power factor lagging. Determine graphically the no-load current of the transformer and its phase with respect to the voltage.

Solution-6

\[\begin{aligned} K & =200 / 800=1 / 4 \\ I_2 & = 80\angle -\cos^{-1}(0.8) \\ I_1 & = 25\angle -\cos^{-1}(0.707)\\ I_0 & = I_1-I_2^{\prime}\\ & = I_1-KI_2\\ & = 5.92\angle-73.57^{\circ} \end{aligned}\]

Problem-7

A single-phase transformer takes 10 A on no-load at power factor of 0.2 lagging. The turns ratio is 4:1 (step down). If the load on the secondary is 200 A at a power factor of 0.85 lagging. Find the primary current and power factor. Neglect the voltage-drop in the winding.

Solution-7

\[\begin{aligned} I_{0} &=10 \angle-78.5^{\circ} \\ &=2-j 9.8 \text { amp } \\ I_{2}^{\prime}&=(200/4) \angle-31.8^{\circ} \\ I_{2}^{\prime}&=42.5-j 26.35 \\ I_{1} & =I_{0}+I_{2}^{\prime} \\ &=2-j 9.8+42.5-j 26.35 \\ &=44.5-j 36.15 \\ &=57.33\angle-39.08^{\circ} \end{aligned}\]