ASSIGNMENTChapter 1

1. What is the effect of temperature on resistance?2. Why do the conductors have positive temperature coefficient? Why do the insulators have negative temperature coefficient?3. What is superconductivity?4. State KCL & KVL and explain it with example.5. Explain the star – delta combination and derive the necessary equations.6. Compare the series and parallel circuit of resistors.7. Derive the equation of the temperature coefficient for two different temperatures:-

Solve the following problems :-1.1 Determine the current supplied by the source in the circuit of fig. 1.11.2 Three resistors are connected in series across a 12 – V battery. The first resistance has a value of 1 ohm, second has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of circuit current.1.3 Find the equivalent resistance between the terminal A and B of the circuit shown in fig. 1.3. Also find the value of currents I1, I2, I3.1.4 In the network shown in fig. 1.4, determine the resistance between A and B. the numbers present the respective resistance in ohm.1.5 Calculate the current flowing through the 10 ohm resistor of fig. 1.5 by using star – delta transformation. All resistances are in ohms.1.6 With the aid of the delta – star transformation, determine the current supplied by the source for a circuit shown in fig. 1.6. all resistance values are in ohms.1.7 Find the resistance between terminals A and B of the circuit shown in fig. 1.7. all resistances are in ohms.1.8 Determine the resistance between the terminal AB of the network shown in fig 1.8.1.9 The unbalanced bridge circuit is shown in fig 1.9. find the value of current in RL = 50 ohm using star – delta transformation.1.10 Find the current in 17 ohm resistor in the network shown in fig 1.10 using star – delta conversion. The numbers indicate the resistance of each member in ohms.

ASSIGNMENTChapter 2

1. Derive the expression of capacitance of a parallel plate capacitor for three different dielectric materials.2. Derive an expression for the equivalent capacitance of a number of capacitors connected in (a) series and (b) parallel.3. Explain the charging and discharging phenomenon of a capacitor through a resistor.4. Why a series resistance is connected during charging and discharging?5. Derive from basic principles, an expression for the energy in a charged capacitor and state where this energy is stored.

SOLVE THE FOLLOWING PROBLEMS:

2.1 Two capacitors of 1μF and 2μF are connected in parallel. A third capacitor of 12μF is connected in series with these two in parallel. Calculate the total capacitance. If a 100 V battery is connected across the whole circuit; calculate the voltage across each capacitor and the charge held by each capacitor.

2.2 Three capacitors of values 2μF, 8μF and 10μF are connected across a 200 V dc supply in parallel. Calculate: a) the resultant capacitance, b) the total charge, c) the charge on each capacitor.

2.3 A 2μF capacitor is connected by closing a switch to a supply of 100 V through 1 MΩ series resistance. Calculate: a) the time constant, b) initial charging current, c) the initial rate of rise of p.d. across capacitor, d) voltage across capacitor 6 seconds after the switch has been closed ande) the time taken for the capacitor to be fully charged.

2.4 A resistance R and a 2μF capacitor are connected in series across a 200 V dc supply. Across the capacitor is connected a neon lamp that strikes at 120 V. Calculate the value of R to make the lamp strike 5 seconds after switch is closed.

2.5 Three capacitors of 10, 20 and 40 μF are placed in series across a 350 V source. Determine: a) charge on each capacitor, b) voltage drop across each capacitor, d) total energy stored in the combination.

2.6 A capacitor of 0.1 μF is charged from a 200 V battery through a series resistance of 1000 Ω. Calculate : a) the time for capacitor to receive 63.2% of its final charge; b) the charge received in this time; c) the initial rate of charging; d) the rate of charging when the charge is 63.2% of its final value.

2.7 A capacitor is made of two plates with an area of 11 cm2 andseparated by a mica sheet of 2mm thick. If for mica, the relative permittivity is 6, find its capacitance. If one of the plates is moved further to give an air gap of 0.5 mm between the plates and mica, find the change in capacitance.

2.8 A capacitor is charged by a dc source through a resistor of 1 MΩ. Ifin one second, the pd across the capacitor reaches 80% of its initial value; calculate its capacity.

2.9 An 8μF capacitor is being charged by a 400 V supply through 0.1 MΩ resistor. How long will it take the capacitor to develop a p.d. of 300 V? Also what fraction of the final energy is stored in the capacitor?

2.10 A capacitor of 80 μF in series with a voltmeter of 10kΩ resistor issuddenly connected across a 100 V d. c. supply. Calculate: i) reading of voltmeter after 0.4 sec. ii) voltage across capacitor after 0.4 sec. iii) the value of current when the voltmeter reads 40 V. iv) the time after which the current becomes equal to 4 mA and v) and charge after 0.2 sec.

ASSIGNMENTCHAPTER 3

1. Define Magnetic circuit. Derive the equation of Flux in terms of M. M. F. and reluctant from first principles. Also explain series and parallel magnetic circuits.2. Give comparison between electrical and magnetic circuits.3. Explain magnetic hysteresis in detail.4. Explain Lenz’s law and Faraday’s law of electro – magnetic Induction.5. Explain statically induced emf.6. Distinguish between (i) dynamically induced emf (ii) statically induced emf.7. Derive expression of equivalent inductance when two magnetically coupled coils are connected in series in two different ways.8. Derive equivalent inductance of any two coupled coils having self inductance L1 and L2, and mutual inductance M when connected in parallel.9. When a d. c. supply of V volts is connected to an inductive circuit having a resistance R and an inductance L through a switch prove that the current in the circuit at any time from the instant of closing the switch is given by

i = V/R (1 – e-Rt/L)10. Write short notes on: 1) coefficient of coupling in coupled inductors.2) Eddy current loss3) Lifting power of magnet4) Force between two parallel current carryingconductors.

SOLVE THE FOLLOWING PROBLEMS:-

3.1 The mean periphery of the steel ring is 50 cm and the cross section area is 4 cm2. Calculate the ampere turns necessary to produce flux of 0.6 mWh. If a saw cut of 2mm is made in the ring and if the mmf remains constant, calculate the new value of flux. Take μr of steel as 1200.

3.2 A series magnetic circuit has an iron path of length 50 cm and an air gap of length 1 mm. the cross sectional area of the iron is 6.66 cm2 and the exciting coil has 400 turns. Determine the current required to produce a flux of 0.9 mWb in the air gap. The following points are taken from the magnetization curve for the iron:

Flux density (T) 1.2 1.35 1.45 1.55Magnetizing force (AT/m) 500 1000 2000 4000

3.3 An iron ring of cross section 3.5 cm2 and mean diameter 40 cm has a saw cut of 1.5 mm width. It is wound with 750 turns of wire. If the flux in the air gap is 0.5 mWb and the current through the coil is 6.5 A, calculate the relative permeability of iron. Take leakage coefficient = 1.15.

3.4 Coil A and B with 50 and 500 turns respectively are wound side by side on a closed iron circuit of section 50 cm2, and mean length of 1.2 m. Estimate:-(i) Mutual inductance between the coils. (ii) Self inductance of each coil and emf induced in coil A if the current in coil B grows steadily from 0 to 5 Amp in 0.01 sec. Assume μr of iron as 1000.

3.5 The number of turns in a coil is 250. When a current of 2 A flows in this coil, the flux in the coil is 0.3 mWb. When this current is reduced to zero in 2 ms, the voltage induced in a coil lying in the vicinity of coil is 63.75 V. if the coefficient of coupling between the coils is 0.85, find self inductances of the two coils, mutual inductance and number of turns in the second coil.

3.6 A flux of 0.5 mWb is produced by a coil of 900 turns wound on a ring with a current of 3 A in it. Calculate (i) Inductance of the coil. (ii) EMF induced in the coil when a current of 5 A switched off assuming the current to fall to zero in 1 millisecond, (iii) the mutual inductance between the coils if a second coil of 600 turns is uniformly wound over the first coil.

3.7 Two coils A & B when connected in series cumulatively has total inductance of 0.5 H. when they are connected in series differentially, the resultant inductance is 0.2 H. If the coil B has self inductance of 0.15 H. Calculate the self inductance of coil A and the induced emf in coil B. When the rate of decrease of current in series combination is 1000 A/sec.

3.8 Two combined inductances of two coils connected in series is 0.76 H and 0.25 H depending upon relative directions of currents in the coils. If one of the coils have isolated, has a self inductance of 0.15 H then calculate (i) mutual inductance and (ii) coefficient of coupling.

3.9 Flux of 1 mWb is produced when a coil A having 800 turns carries a current of 4 A. The same current when flows through a coil B of 1200 turns produces a flux of 1.5 mWb. The two coils are placed beside such that 80 % of the flux produced by coil A links with coil B. find:

(1) Self inductance of each coil.(2) Mutual inductance between the two coils.(3) Co-efficient of coupling(4) Percentage of flux produced by coil B that links coil A.3.10 A coil A having 800 turns is wound over an iron ring of mean diameter of 200 mm and area of cross – section of 1.5 cm2. Another coil B having 600 turns is wound over this coil. Relative permeability of iron is 750. Calculate;-(1) Self – inductance of each coil.(2) Mutual inductance between two coils.

why do we connect the resistor in series with capacitor in charging?

At the instant of closing the switch, there is no charge on the capacitor and hence no potential difference across it. so, the whole of the applied voltage must across a non-inductive resistor. as a result, initial value of charging current becomes equal to V/R which is maximum possible. Thus, the current will be maixmum and it can damage the capacitor.

why do semiconds and inltrs have negative temp. coefficient?

Actually, semiconductors and insulators have negative temperature coefficient…bcz they have energy gap below 3 eV and above 3 eV respectively. so, as the temperature increases the electrons from the valance band can not jump to the conduction band.secondly, the no. of free electrons in the semicoductors and insulators is only 1, but that goes in the negative once the temperature rises.

important que for mid sem eee

derive the equation of temperature coefficient.explain star delta transformationexplain charging and discharging of capacitorexplain charging of inductorexplain series connection of inductor and derive the equation for the co-efficient of coupling.why do semiconductors have negative temperature co-efficient?why do we connect the resistor in charging circuit of capacitor?

AssignmentChapter 4

1) Draw and explain stair case wiring.2) Draw and explain Godown wiring3) Draw and explain High-rise building wiring4) What is fuse? why is it required? what are the properties of fuse element?

AssignmentChapter 5

1) Derive relationship of phase voltage & line voltage, Phase current & line current in case of star connections and delta connections in 3-phase circuit.

2) State the Advantages of 3-phase system over 1-phase system.3) Explain types of methods for measurement of power in case of 3 phase circuits. Derive the equation of power in both delta and star connection.4) Explain the method of measuring power in 3 phase circuit by two wattmeter method. What are the effects of different power factor?