Subject: BASIC ELECTRICAL ENGG. Subject Code: BE2102 2ND Semester: Applicable Branch (es): ALL Question Code: F-463 Year of Examination:2014 Exam Type: (Special/Regular): REGULAR Solution Prepared by Sl.No. Faculty Name Email ID 1 Mr. CH. MURTHY firstname.lastname@example.org 2 Ms. SASMITA PADHY email@example.com 3 Mr. B.RAJANARAYAN PRUSTY firstname.lastname@example.org 4 Ms. G.SIVARANJANI email@example.com 5 Mr. Mitu Baral firstname.lastname@example.org NATIONAL INSTITUTE OF SCIENCE &TECHNOLOGY PALUR HILLS, BERHAMPUR, ORISSA – 761008, INDIA
1. ANSWER ALL THE QUESTIONS [10X2=20] (a) Explain how voltage source with a source resistance can be converted in to an equivalent current source? Ans: Sometimes, while analyzing a circuit, it is convenient to replace a voltage source with an equivalent current source and vice versa. This is simply a mathematical conversion on paper but not a physical conversion. Two sources are equivalent if, for any load resistor connected to the two sources, they produce the same voltage across that resistor and the same current through it. Figure (a) and (b) below shows the practical voltage source and its equivalent current source. (b) State the advantages of sinusoidal alternating quantity? Ans: 1. Mathematically it is very easy to write the equations for purely sinusoidal wave from. Any other type of waveform can be resolved in to a series of cosine waves of fundamental and higher frequencies. Sum of all these waves give the original wave from. 2. The sine and cosine waves are the only waves which can pass through linear circuits containing resistance, inductance and capacitance. In case of other waveforms, there is a possibility of distortion when it passes through linear circuit. 3. The integration and derivative of a sinusoidal function is again a sinusoidal function. This makes the analysis of linear electrical networks with sinusoidal input, very easy (c) List any two advantages of 3-phase system over 1-phase system? Ans: (1) The weight of the conductors and other components in a three-phase system is much lower than in a single phase delivering the same amount of power. (2) Power produced by a single phase system has a pulsating nature, where as three phase system can deliver a steady, constant supply of power. (3) Three phase motors have a non-zero starting torque, unlike their single phase counter part (d) Does a transformer draw any current when secondary part is open? Why? Ans: Transformer draws some current even when secondary is open to supply iron loss. (Hysteresis loss and eddy current loss)
Let I0=no load primary current. I0 have two components IW and IM IW supplies the iron loss and primary copper loss at no load = I0sinø IM- magnetizing component. Its function is to sustain the alternating flux in the core and it is wattles = I0cosø (e) Why the armature core in d.c machines is constructed with laminated steel sheets instead of solid steel sheets? Ans: To reduce eddy current loss. To work efficiently avoiding unnecessary overheating of the machine core are laminated to reduce circulating currents, called eddy currents, which are caused by voltages induced in to the core. (f) What is the function of a capacitor in a single phase induction motor? Ans: It provides the phase shift necessary to have a rotating magnetic field for the rotor to spin. It also improves the power factor. (g) A stereo amplifier is providing 50watts of output power. How much power input is required if the amplifier is 30%efficient. Ans: Efficiency, η = Output power/Input power …………………….(1) Given: Efficiency, η = 30%; Output power = 50 watts; From (1) Input power = 166.7 watts. (h) What do you mean by magnetism and magnetic field? Ans: Magnetism: Magnetism refers to physical phenomena arising from the force between magnets, objects that produce fields that attract or repel other objects. All materials experience magnetism, some more strongly than others. Permanent magnets, made from materials such as iron, experience the strongest effects, known as ferromagnetism. Then there's paramagnetism, in which certain materials are attracted by a magnetic field, and diamagnetism, in which materials are repelled by a magnetic field. Other, more complex, forms include anti-ferromagnetism, in which the magnetic properties of atoms or molecules align next to each other; and spin glass behavior, which involve both ferromagnetic and anti-ferromagnetic interactions. Some materials are called non-magnetic, because their magnetic effects are so small. Magnetism can also vary depending on temperature and other factors. Magnetic field: A magnetic field is a way of mathematically describing how magnetic materials and electric currents interact. Magnetic fields have both a direction and a magnitude, or strength. Magnets have a "north" pole and a "south" pole. Opposite poles attract each other and alike poles repel each other. These poles are referred to as a magnetic dipole. Magnetic dipoles and electric currents both give rise to magnetic fields.
(i) Why do electric lines of force never cross? Ans: The electric lines of force are imaginary lines. A unit positive charge placed in the electric field tends to follow a path along the field line if it is free to do so. The electric lines of force emanate from a positive charge and terminate on a negative charge. The tangent to an electric field line at any point gives the direction of the electric field at that point. Two electric lines of force can never cross each other. If they do, then at the point of intersection, there will be two tangents. It means there are two values of the electric field at that point, which is not possible. Further, electric field being a vector quantity, there can be only one resultant field at the given point, represented by one tangent at the given point for the given line of force. Electric lines of force are closer (crowded) where the electric field is stronger and the lines spreadout where the electric field is weaker. (j) If a capacitor of 0.05mfd in series circuit provides a resonant frequency of 833 kHz, what is the value of inductance? Ans: Given: C=0.05 micro farad fr=833kHz At resonance condition, XL=XC ωL=1/ωC ω2=1/LC L=1/ω2C=7.3*10-7 Henry 2. (a) A mild steel ring has a mean diameter of 160 mm and cross section area of 300 mm 2. Calculate the mmf required to produce a flux of 333 micro webers, reluctance and relative permeability.  The B-H data is given below: B(T) H(AT/m) 0.9 260 1.1 450 Ans: Given: d=160*10-3m A=300*10-6m2 φ=3.33*10-6wb MMF= ? Reluctance, S= ? µr= ? As, B=µH and φ = A*B B= φ/A =333*10-6/300*10-6= 1.11 wb/m2 From the table given above, 1.2 600 1.3 820