Circuit and Working Principle of Autotransformer StarterThe circuit diagram of an autotransformer starter for starting a 3-phase induction motor is shown in the figure. The autotransformer starter can be used for starting both star and delta connected 3-phase induction motors. In this method, the starting current of the motor is limited by using a 3-phase autotransformer to decrease the initial applied voltage to the stator. The autotransformer is provided with a number of tappings to obtain the variable voltage.In the autotransformer starting method, the starter is connected to a particular tapping of the autotransformer to obtain the most suitable starting ... Read More

Applications of 3-Phase Wound-Rotor Induction MotorThe slip-ring or wound-rotor 3-phase induction motors are used in the following applications −Slip ring induction motors are suitable for loads requiring high starting torque and for applications where the starting current is low.Slip ring induction motors are used for loads having high inertia, which results in very high rotor energy losses during acceleration.The slip ring induction motors are also used for loads which require a gradual build-up of load.They are used for loads that requires speed control.Typical applications of wound rotor or slip ring induction motors are crushers, plunger pumps, cranes & hoists, elevators, ... Read More

Rotor Current FrequencyThe frequency of current and voltage in the stator of a 3-phase induction motor must be same as the supply frequency and is given by, $$\mathrm{𝑓 =\frac{𝑁_{𝑆}𝑃}{120}… (1)}$$But, the frequency of the current and EMF in the rotor circuit of the 3-phase induction motor is variable and depends upon the difference between the synchronous speed (NS) and the rotor speed (Nr), i.e., on the slip. Thus, the rotor frequency is given by, $$\mathrm{𝑓_{𝑟} =\frac{(𝑁_{𝑆} − 𝑁_{𝑟} )𝑃}{120}… (2)}$$Now, from the equations (1) and (2), we get, $$\mathrm{\frac{𝑓_{𝑟}}{𝑓}=\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}}$$$$\mathrm{∵ \:Slip, \:𝑠 =\frac{𝑁_{𝑆} − 𝑁_{𝑟}}{𝑁_{𝑆}}}$$$$\mathrm{∴ 𝑓_{𝑟} = 𝑠𝑓 … ... Read More

The torque is defined as the turning moment of a force about an axis. It is measure by the product of the force (F) and perpendicular distance (r) of the line of action of force from the axis of rotation, i.e., $$\mathrm{Torque, \: 𝜏 = 𝐹 × 𝑟 \:… (1)}$$The torque is measured in Newton-meters (Nm).Armature Torque of DC MotorIn a DC motor, a circumferential force (F) at a distance r which is the radius of the armature is acted on each conductor, tending to rotate the armature. The sum of the torques due to all the armature conductors is ... Read More

Swinburne’s test is an indirect method of testing DC machines, named after Sir James Swinburne. In this method, the losses are determined separately and the efficiency at desired load is predetermined. The Swinburne’s test is the simplest method of testing of shunt and compound DC machines which have constant field flux.The connection diagram is shown in the figure and the machine is run as a motor at rated voltage and speed.Let, $$\mathrm{𝑉 = Supply\:voltage}$$$$\mathrm{𝐼_{0} = No \:load \:line\: current}$$$$\mathrm{𝐼_{sh} = Shunt\: field \:current}$$$$\mathrm{\therefore \:No \:load\: armature \:current, \:I_{𝑎0} = I_{0} − I_{sh}}$$And$$\mathrm{No load input power = 𝑉𝐼_{0}}$$This no-load input power ... Read More

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