Found 1006 Articles for Electronics & Electrical

The speed of a DC shunt is given by, $$\mathrm{𝑁 \varpropto\frac{𝐸_{𝑏}}{\varphi}}$$$$\mathrm{⇒ 𝑁 = 𝐾 (\frac{𝑉 − 𝐼_{𝑎}𝑅_{𝑎}}{\varphi})\: … (1)}$$It is clear from the equation (1) that the speed of a DC shunt motor can be changed by two methods −Flux Control MethodArmature Resistance Control MethodFlux Control MethodThe flux control method is based on the principle that by varying the field flux ϕ, the speed of DC shunt motor can be changed.$$\mathrm{𝑁 \varpropto\frac{1}{\varphi}}$$In this method, a variable resistance (called field rheostat) is connected in series with the shunt field winding. By increasing the resistance of the field rheostat, the shunt field ... Read More

The speed of a DC series motor is given by, $$\mathrm{𝑁 \varpropto\frac{𝐸_{𝑏}}{\varphi}}$$$$\mathrm{⇒ 𝑁 = 𝐾 (\frac{𝑉 − 𝐼_{𝑎}(𝑅_{𝑎} + 𝑅_{𝑠𝑒})}{\varphi}) … (1)}$$Hence, it is clear from the eq. (1) that the speed of a DC series motor can be changed by using any one of the following two methods −Field Control MethodArmature Resistance Control MethodField Control MethodThe field control method is based on the fact that by varying the field flux in the series motor, its speed can be changed, as, $$\mathrm{N \varpropto\frac{1}{\varphi}}$$The change in the flux can be achieved by in the following ways −Field DiverterIn this method, a ... Read More

Speed of a DC MotorThe expression for the speed of a DC motor can derived as follows −The back EMF of a DC motor is given by, $$\mathrm{E_{b} = V − I_{a}R_{a} … (1)}$$Also, $$\mathrm{𝐸_{b} =\frac{NP\varphi𝑍}{60𝐴}\:… (2)}$$From eq. (1) & (2), we get, $$\mathrm{\frac{NP\varphi Z}{60A}= V − I_{a}R_{a}}$$$$\mathrm{⇒ N = (\frac{V − I_{a}R_{a}}{\varphi}) \times\frac{60A}{PZ}}$$For a given DC motor, the (60A/PZ) = K (say) is a constant.$$\mathrm{\therefore N = K (\frac{V − I_{a}R_{a}}{\varphi})}$$But, $$\mathrm{(V − I_{a}R_{a}) = E_{b}}$$Therefore, $$\mathrm{N = K (\frac{𝐸_{𝑏}}{\varphi}) \:… (3)}$$$$\mathrm{⇒ N \varpropto\frac{E_{b}}{\varphi}\:......(4)}$$Hence, the speed of a DC motor is directly proportional to back emf and is inversely ... Read More

As in a practical transformer, the no-load current I0 is very small as compared to rated primary current, thus the drops in R1 and X1 due to the I0 can be neglected. Therefore, the parallel circuit R0 – Xm can be transferred to the input terminals. The figure shows the simplified equivalent circuit of the transformer.The simplified equivalent circuit can be referred to primary side or secondary as discussed below (here, the assumed transformer is step-up transformer).simplified Equivalent Circuit Referred to primary SideThis can be obtained by referring all the secondary side quantities to the primary side as shown in ... Read More

When the secondary winding of a practical transformer is open circuited, the transformer is said to be on no-load (see the figure). Under this condition, the primary winding will draw a small no-load current I0 from the source, which supplies the iron losses and a very small amount of copper loss in the core and primary winding respectively. Thus, the primary no-load current (I0) does not lag the applied voltage V1 by 90° but lags it by an angle φ0 which is less than 90°.Therefore, $$\mathrm{N_{o} − load\:input\: power, \:𝑃_{0} = 𝑉_{1}𝐼_{0} cos\varphi_{0}}$$From the phasor diagram, it can be seen ... Read More

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