Conquering Transformer Basics: Solved Problems for Smooth Sailing

Demonstrative Video

Problem-1

The maximum flux density in the core of a 250/3000 Volts, 50 Hz single phase transformer is 1.2 Wb/m\(^2\). If the emf per turn is 8 volt. Determine:

  1. primary and secondary turns

  2. Area of the core

Solution-1

\[\begin{aligned} E_{1} &=N_{1} \times \text{e.m.f. induced/turn} \\ N_{1} &=250 / 8=32 \\ N_{2}&=3000 / 8=375 \\ E_{2} &=4.44 f N_{2} B_{\text {m }} A \\ 3000 &=4.44 \times 50 \times 375 \times 1.2 \times A \\ A&=0.03 \mathrm{~m}^{2} . \end{aligned}\]

Problem-2

The core of a 100-KVA, 11000/550 V, 50-Hz, 1-ph, core type transformer has a cross-section of 20 cm \(\times\) 20 cm. Find

Solution-2

\[\begin{aligned} B_{m}&=1.3 T \\ A&=(0.2 \times 0.2) \times 0.9=0.036 \mathrm{~m}^{2}\\ 11,000 &=4.44 \times 50 \times N_{1} \times 1.3 \times 0.036 \\ \Rightarrow N_{1}&=1060 \\ 550 &=4.44 \times 50 \times N_{2} \times 1.3 \times 0.036 \\ N_{2}&=53 \\ N_{2} &=K N_{1}=(550 / 11,000) \times 1060=53\\ \mathrm{emf/tum}&=11,000 / 1060=10.4 \mathrm{~V} \text { or } 550 / 53=10.4 \mathrm{~V} \end{aligned}\]

Problem-3

A single-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm\(^2\). If the primary winding be connected to a 50-Hz supply at 520 V, calculate

  1. the peak value of flux density in the core

  2. the voltage induced in the secondary winding.

Solution-3

\[\begin{aligned} K &=N_{2} / N_{1}=1000 / 400=2.5 \\ E_{2} / E_{1} &=K \\ E_{2}&=K E_{1}=2.5 \times 520=1300 \mathrm{~V} \\ E_{1} &=4.44 f N_{1} B_{m} A \\ 520 &=4.44 \times 50 \times 400 \times B_{m} \times\left(60 \times 10^{-4}\right)\\ B_{m}&=0.976 \mathrm{~Wb} / \mathrm{m}^{2} \end{aligned}\]

Problem-4

A 25 KVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to a 3000-V, 50-Hz supply. Find

Neglect leakage drops and no-load primary current.

Solution-4

\[\begin{aligned} K&=N_{2} / N_{1}=50 / 500=1 / 10 \\ I_{1} & =25,000 / 3000=8.33 \mathrm{~A} \\ I_{2}&=I_{1} / K=10 \times 8.33=83.3 \mathrm{~A} \\ E_1~\text { per turn } & =3000 / 500=6 \mathrm{~V} \\ E_2 & =6 \times 50=300 \mathrm{~V} \\ &\text { (or } E_{2}=K E_{1}=3000 \times 1 / 10=300 \mathrm{~V} \text { ) } \\ E_{1}&=4.44 \mathrm{fN}_{1} \Phi_{\mathrm{m}} \\ \Rightarrow 3000&=4.44 \times 50 \times 500 \times \Phi_{\mathrm{m}} \\ \Rightarrow \Phi_{\mathrm{m}}&=27 \mathrm{mWb} \end{aligned}\]

Problem-5

The core of a three phase, 50 Hz, 11000/550 V delta/star, 300 KVA, core type transformer operates with a flux of 0.05 Wb. Find

  1. number of H.V. and L.V turns per phase

  2. emf per turn

  3. full load HV and LV phase currents.

Solution-5

\[\begin{aligned} \phi_m & = 0.05 \mathrm{~Wb}\\ \mathrm{EMF/turn} & = 4.44 \mathrm{f} \phi_{m} \\ &=4.44 \times 50 \times 0.05=11.1 \mathrm{volts} \\ \text{Voltage per phase on delta side} &=11000~\text{V}\\ \text{Voltage per phase on star side} &=550 / 1.732=317.5~\text{V} \\ T_{1} & = \text{ voltage per phase/e.m.f. per turn}\\ &=11000 / 11.1=991 \\ T_{2}&= 317.5 / 11.1=28.6\\ \text { Output per phase }&=(300 / 3)=100 \mathrm{kVA} \\ \text { H. V. phase-current }&=\frac{100 \times 1000}{11,000}=9.1 \mathrm{Amp} \\ \text { L. V, phase-current }&=(100 \times 1000 / 317.5)=315 \mathrm{Amp} \end{aligned}\]

Problem-6

A single-phase transformer has 500 turns in the primary and 1200 turns in the secondary. The cross-sectional area of the core is 80 sq.cm. If the primary winding is connected to a 50-Hz supply at 500 V, calculate

  1. Peak flux-density, and

  2. Voltage induced in the secondary

Solution-6

\[\begin{aligned} E_1 & = 500 =4.44 \times 50 \times \phi_{m} \times 500 \\ \Rightarrow \phi_{m} &=4.50 \mathrm{~mWb} \\ B_{m} &=\phi_{m} /\left(80 \times 10^{-4}\right)=0.563 \mathrm{wb} / \mathrm{m}^{2}\\ \frac{V_{2}}{V_{1}} &=\frac{N_{2}}{N_{1}} \\ V_{2} &=500 \times 1200 / 500=1200 \text { volts } \end{aligned}\]

Problem-7

A 25-KVA, single phase transformer has 250 turns on the primary and 40 turns on the secondary winding. The primary is connected to a 1500-volt, 50-Hz mains. Calculate

Solution-7

\[\begin{aligned} \dfrac{V_2}{1500} & = \dfrac{N_2}{N_1}=\dfrac{40}{250}\\ \Rightarrow V_2 & = 240~\mathrm{Volts} \\ I_1 & = 25000/1500 = 16.67~\mathrm{A} \\ I_2 & = 25000/240 = 104.2~\mathrm{A} \\ E_1 & = 1500 = 4.44\times 50\times \phi_m\times 250\\ \Rightarrow \phi_m & = 0.027~\mathrm{Wb} \end{aligned}\]

Problem-8

An ideal transformer has a 150 turn primary and 750 turn secondary. The primary is connected to a 240-V, 50-Hz source. The secondary winding supplies a load of 4 A at a lagging power factor of 0.8. Determine:

Solution-8

\[\begin{aligned} a & =150 / 750=0.2 \\ I_{2}&=4 \mathrm{~A} \\ I_{1}&=\frac{I_{2}}{a}=\frac{4}{0.2}=20 \mathrm{~A} \\ V_{2}&=\frac{V_{1}}{a}=\frac{240}{0.2}=1200 \mathrm{~V} \\ P_{L} &=V_{2} I_{2} \cos \theta=1200 \times 4 \times 0.8=3840 \mathrm{~W}\\ \Phi_{m} &=\frac{E_{1}}{4.44 f N_{1}}=\frac{V_{1}}{4.44 f N_{1}}=\frac{240}{4.44 \times 50 \times 150} \\ &=7.21 \mathrm{mWb} \end{aligned}\]