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**Chapter 1:**

Generation , transmission, distribution and utilization of electrical energy and its importance.

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**Generation of Alternating EMF**

Consider a rectangular coil of N turns rotating in a uniform magnetic field with an angular velocity ω radian/second about X-axis in its own plain as shown in fig.3.6. Let фm be the maximum flux perpendicular to the axis of rotation, when the plain of the coil coincides with the X-axis.

**Faraday’s Laws of Electromagnetic Induction**

It is well known that whenever an electric current flows through a conductor, a magnetic field is generated around the conductor. And the converse of this is also true i.e. when a magnetic field around a conductor moves relative to the conductor, it produce a flow of electrons in the conductor. This phenomenon whereby an emf and hence current is induce in any conductor is cut by a magnetic flux is known as electromagnetic induction.

**Maxwell’s Loop Current Method (Mesh Analysis)**

This method is particularly well-suited to coupled circuit solutions employs of loop or mesh current instead of branch currents as in Kirchhoff’s laws. In this method the currents in different mesh are assigned continuous paths so that they do not split at a junction into branch currents. Basically, this method consists of writing loop voltage equation by Kirchhoff’s voltage law in terms of unknown loop currents.

**Electric Circuit and Circuit Element**

As we know that the electric current is the rate of flow of electric charge (electrons). It is denoted by I and its unit is Ampere (A) and a complete path which consist one or more circuit elements and connecting wire and allows flowing electric current is known as electric circuit. Read more »

EMF Equation

Derive the emf equation of a DC generator.

E= dΦ/dt = ZΦN/60 x P/A Volt

Where, Φ = flux/pole in wb; Z = total number of armature conductors = No.of slots x No.of conductors/slot; P = No.of generator poles; A = No.of parallel paths in armature; N = armature rotation in revolutions per; minute (RPM); E = e.m.f induced in any parallel path in armature

**Power And Power Factor**

The component of current which is in phase with the applied voltage (I cosΦ) and contributes to active or true power of the circuit is known as active or watt-ful or in phase component of alternating current

and the component of current which is perpendicular to applied voltage (I sinΦ)and contributes to reactive power of the circuit is known as reactive or watt-less component of alternating current.

**Analogies between magnetic circuits and electrical circuits**

Analogies between magnetic circuits and electrical circuits are easy to be made by comparing electric circuit and magnetic circuit.

§ The magnetic flux Φ which flows( not actually flow but sets up) through a magnetic circuit, corresponds to the electrical current / which flows through an electrical circuit;

**Sample Question:**

A 4-pole dc shunt generator with a shunt field resistance of 100ohm and an armature resistance of 1 ohm has 378 wave connected conductors in its armature. the flux per pole is 0.02wb. If a load resistance of 10 ohm is connected across the armature terminals and generator is driven at 1000 RPM, calculate the power absorbed by the load.

**Solution:**

We know that from the reference note Click here to view Emf generated in armature

E= ZΦN/60 x P/A Volt

Here, Φ = 0.02wb; P =4; N = 1000; A=2 for wave winding then

E= 252 Volt

Armature current Ia = Generated emf / total resistance = E/ ((10//100)+1) =252Volt /9.09 ohm=27.72A

( load resistance and shunt resistance are parallel then armature resistance is in series with this combination)

Terminal Voltage V= E- IaRa =224.27 V

Power absorbed by Load PL= V2 / RL = 224.272 / 10 =5029.7 Watt Ans.