When a load Impedance is connected across the secondary winding of the practical transformer, then the transformer is said to be loaded and draws a load which flows through the secondary winding and the load.We shall consider following two cases for analysing the practical transformer −Case 1 – When the transformer is assumed to have no winding resistance and leakage fluxThe figure shows a practical transformer with the assumption that the winding resistances and the leakage reactances are neglected. With this assumption, $$\mathrm{𝑉_{1} = 𝐸_{1}\: and \:V_{2} = 𝐸_{2}}$$Consider an inductive load is connected across the secondary winding which causes ... Read More

Hopkinson’s test is a method of testing the efficiency of DC machines. The Hopkinson’s test is known as regenerative test or back-to-back test or heat-run test.This test requires two identical shunt machines which are mechanically coupled and also connected electrically in parallel. One machine acts as a motor and the other as a generator.The motor takes its input from the supply and the mechanical output of the motor drives the generator. The electrical output of the generator is used in supplying the input to the motor. Therefore, the output of each machine is fed as input to the other.When both ... Read More

No-Load Equivalent Circuit of TransformerThe figure shows the no-load equivalent circuit of a practical transformer. In this, the practical transformer is replaced by an ideal transformer with a resistance R0 and an inductive reactance Xm in parallel with its primary winding. The resistance R0 represents the iron losses so the current IW passes it and supplies the iron losses. The inductive reactance Xm draws the magnetising current Im which produces the magnetic flux in the core.Therefore, $$\mathrm{Iron\:losses \:of\: practical\: transformer\: = I_W^2 𝑅_{0} =\frac{𝑉_{1}^2}{𝑅_{0}}}$$Also, from the equivalent circuit, $$\mathrm{𝑉_{1} = 𝐼_{𝑊}𝑅_{0} = 𝐼_{𝑚}𝑋_{𝑚}}$$The no-load current is given by phasor sum ... Read More

EMF Equation of TransformerConsider an alternating voltage is applied across the primary winding of the transformer and the frequency of the supply voltage is f. This applied voltage produces a sinusoidal flux φ in the core of the transformer, which is given by, $$\mathrm{\varphi = \varphi_{𝑚} sin \omega t}$$Due to this sinusoidal flux, an EMF is induced in the primary winding, whose instantaneous value is given by, $$\mathrm{𝑒_{1} = −𝑁_{1}\frac{𝑑\varphi}{𝑑𝑡} = −𝑁_{1}\frac{𝑑}{𝑑𝑡}(\varphi_{𝑚} sin \omega t)}$$$$\mathrm{⇒ 𝑒_{1} = −𝑁_{1}\omega\varphi_{𝑚} cos \omega t = 𝑁_{1}\omega\varphi_{𝑚} sin (\omega t −\frac{\pi}{2})}$$$$\mathrm{\because \: \omega = 2\pi𝑓}$$$$\mathrm{\therefore 𝑒_{1} = 2𝜋𝑓𝑁_{1}\varphi_{𝑚} sin (\omega t −\frac{\pi}{2})}$$$$\mathrm{⇒ 𝑒_{1} = ... Read More

When the armature of a DC generator rotates in magnetic field, an emf is induced in the armature winding, this induced emf is known as generated emf. It is denoted by Eg.Derivation of EMF Equation of DC GeneratorLet$$\mathrm{\varphi = Magnetic\: flux \:per\: pole\: in\: Wb}$$$$\mathrm{𝑍 = Total \:number \:of \:armature\: conductors}$$$$\mathrm{𝑃 = Number \:of \:poles\: in\: the \:machine}$$$$\mathrm{𝐴 = Number\: of \:parallel \:paths}$$$$\mathrm{Where, \:𝐴 = 𝑃\: … for \:LAP \:Winding \:= 2\: … for \:Wave \:Winding}$$$$\mathrm{𝑁 = Speed \:of \:armature\: in \:RPM}$$$$\mathrm{E_{g} = Generated \:EMF = EMF \:per\: parallel \:path}$$Therefore, the magnetic flux cut by one conductor in one revolution ... Read More

The full-load iron losses and copper losses can be determined by open-circuit test and short-circuit test respectively. Therefore, From the open circuit test, $$\mathrm{Full \:load\: iron \:loss \:= \:P_{i} Watt}$$From the short-circuit test, $$\mathrm{Full \:load\: copper \:loss = P_{cu} Watt}$$$$\mathrm{\therefore\:Total\: full\: load \:losses\: = P_{i} + P_{cu} Watt}$$Therefore, the efficiency of transformer at full-load is$$\mathrm{\eta_{fl} =\frac{VI_{fi} × cos\varphi2}{(VI_{fi} × cos \varphi2) + P_{i} + P_{cu}}… (2)}$$Now, for fraction of load, i.e., if the load is x &tImes; Full load, then, $$\mathrm{Corresponding\: total \:losses = P_{i} + x^2P_{cu}}$$The iron loss (Pi) remains constant at all loads.Thus, the efficiency corresponding to fraction ... Read More

The util.promisify() method basically takes a function as an input that follows the common Node.js callback style, i.e., with a (err, value) and returns a version of the same that returns a promise instead of a callback.Syntaxutil.promisify(original)ParametersIt takes only one parameter −original −This parameter takes an input for the function and returns them as a promise.Example 1Create a file with the name "promisify.js" and copy the following code snippet. After creating the file, use the command "node promisify.js" to run this code.// util.promisify() Demo example // Importing the fs and util modules const fs = require('fs'); const util = ... Read More

An ideal transformer is an imaginary transformer which has the following characteristics −The primary and secondary windings have negligible (or zero) resistance.No leakage flux, i.e., whole of the flux is confined to the magnetic circuit.The magnetic core has infinite permeability, thus negligible mmf is require to establish flux in the core.There are no losses due winding resistances, hysteresis and eddy currents. Hence, the efficiency is 100 %.Woking of Ideal TransformerIdeal Transformer on No-LoadConsider an ideal transformer on no-load, i.e., its secondary winding is open circuited (see the figure). Thus, the primary winding is a coil of pure inductance.When an alternating ... Read More

Transformer A transformer is a static electrical machine which is used for either increasing or decreasing the voltage level of the AC supply with a corresponding decrease or increase in the current at constant frequency. Parts of a Transformer A typical transformer essentially consists of following main parts − Magnetic Core Transformer Windings Transformer Body / Tank and Dielectric Oil Oil Conservator Tank Breather Magnetic Core The core of the transformer is made up of magnetic materials having high permeability. As the transformer is subjected to the ... Read More

A practical transformer is the one which has following properties −The primary and secondary windings have finite resistance.There is a leakage flux, i.e., whole of the flux is not confined to the magnetic circuit.The magnetic core has finite permeability, thus a considerable amount of mmf is require to establish flux in the core.There are losses in the transformer due to winding resistances, hysteresis and eddy currents. Therefore, the efficiency of a practical transformer is less than 100 %.The figure shows a typical practical transformer, which possess all the characteristics that are described above.Winding ResistancesAs the windings of a transformer are ... Read More