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A condition cannot exist when one is plus and the other minus. Similarly the flux c12 which is linking coil 2 of N 2 turns and is due to current J1 in coil 1. If a core is of good magnetic material, the coefficient of coupling could be as high as 0. The mutual inductance di induces voltage in these coils.
The coil 1 will induce a voltage M a: If these are opposing the main di 1 di2 fuxes then these voltages will be - M d and - M d respectively. Assume that the core is Ni initially unmagnetised. If the current i is increased i. The flux density B starts increasing as shown by oa part of the curve in Fig. The point a corresponds to a particular value of current say i 1.
If i were less than i 1 , point a will shift downwards towards o and a smaller final loop will appear. Now if we decrease magnetic field intensity i. We see that when current through the coil is zero still there is some magnetic flux density in the core of the coil and this is known as residual flux density corresponding to point b in the Fig.
If His now reversed by reversing the fow of current i, the fux in the core will decrease and the point on the B-H curve will move along be and when H corresponding to-H e is reached, the core is again demagnetised. If His made corresponding current - i 1 , the point on the B-H Ni 1 curve moves along the cd path.
If His now decreased to zero and then increased to the value corresponding to current i, the B-H curve will follow the path defa. Usually the last point a is slightly short of the original point but for simplicity we assume that it coincides the original point while moving from d through ef and towards 'a'. However, it is found that, if the core is subjected to a few cycles of magnetisation, the loop almost closes and this loop is called hysteresis loop.
It is to be noted that at point b the core is magnitised even though the current through the coil is zero. Throughout the complete cycle of magnetisation the flux density lags behind the magnetic field intensity. Smaller hysteresis loops can be obtained by varying maximum value of current and hence magnetic field intensity. The corresponding hysteresis loops are shown in Fig.
The locus of the tips of the hysteresis loops shown with dashed line is known as magnetisation curve. If the iron core is magnetised from an initial unmagnetised condition, the flux density will follow the megnetisation curve. When current i is varied over one cycle, during some part of the cycle, the coil stores energy from the source and during some part, it returns energy back to the source.
However, energy flowing into the coil is greater than the energy returned back to the source. Hence, during one cycle of variation of current i Hence H there is a net energy flow from the source to the coil core assembly. The energy difference goes to heat the core. The loss of power in the core due to hysteresis effect is known as hysteresis loss.
Substituting these values in equation ii. Since it is not possible to evaluate the area of the B-H loop mathematically as the curve can't be described by some mathematical expression. Steinmetz suggested empirical relation after lot of experimentation to the area of the loop. Both k and n can be obtained empirically.
Using equation ii. Typical value for grain oriented silicon steel sheet used for electric mechines is 1. Some typical values for k are: These eddy currents result in a loss in magnetic materials known as eddy current loss. The circulation of these currents has been shown in Fig. It has been observed that eddy currents flow cross wise in the core. The magnitude of eddy current can be reduced substantially by increasing the cross-sectional resistance of the core.
The cores are made of the sheets called lamination. Each lamination is insulated from the others by a coating of oxides or varnish or both so that it offers a high resistance to the flow of eddy currents. The eddy current loss can also be reduced by using a high resistivity core material.
The sum of the hysteresis and eddy current loss is known as core loss or iron loss. Since for a machine f and Em are constant, the core loss is also known as constant loss of the machine. Eddy currents can be used to advantage to electric braking purpose e. Also there are some specially designed machines which utilise eddy current braking. Example ii. A wire of length 50 ems moves at right angles to its length at 50 ml sec in a uniform magnetic field of density 1 Wb!
Determine the e. Determine the instantaneous value of the induced e. Hence the angle between normal to the plane and magnetic! An airplane has O. The vertical component of the magnetic flux density of the earth's magnetic field is 50 j1 Determine the emf induced between the wing tips. The e. The end of the magnet is withdrawn through a 5 00 turn coil in 1 I1 0 of a sec.
What is the e.
A circular iron core has cross sectional area of 5 cm2 and length of magnetic path 50 ems. It has two coils A and coil B. Coil A has turns and B turns and the current 1 in coil A is changed from 0 to 1 5 A in J sec. Determine the emf induced in coil B. A conductor 0. The bus bars are arranged 30 ems apart. I1I 2 l 2n r where l is the length of the conductor which we will take use as. Since it is a single phase bus bar system! Since the effective impedance is. Hence force per metre will be.
Since the currents are flowing in opposite direction, there will be force of repulsion.
The flux q linked by a coil of turns varies during the period T of one T T complete cycle as follows: If T be: Since the flux is a time function and the coefficient T is common in the complete duration. V Ans. Calculate the inductance of a toroid 25 cm mean diameter and 6.
Here minus sign as the current is increasing. An air cored coil has turns with an inductance of mH. Determine the new value of inductance. Te current in an ignition coil is 3A. If the voltage across the spark plug is 20 kV determine the current.
The energy stored in the inductor is. There are two coils having coefficient of coupling 0. The current in coil 1 is 3 A and the total flux 0. The voltage induced in coil 2 is 85 volts when the current in coil is reduced to zero in 3 m sec.
The no of turns in coil 1 is Determine L1, L 2 , Mand N2. Jn a pair of coupled coils , coil 1 has turns and carries a current of 2. A current of 5 A in X produces in it a fux of 0. If the current in X changes from 6 A to - 6 A in 1 0 m sec. Since the two coils are identical. Determine the hourly loss of energy in a specimen of iron the hysteresis loop of which is equivalent in area of JI m 3. Frequency 50 Hz, density of iron I5 and weight l O kg.
Density is 7.
The maximum flux density in a hysteresis loop is 1. The area of the hysteresis loop is 0. Determine the hysteresis loss if 1 cm 3 of the material is subjected to an alternating fux density of 1. The area of the loop 0. The hysteresis loss in a sample of iron is found to be 4. Te density of iron 7. Two coils with N1 and N2 turns are wound on the same magnetic bar. State and explain Faraday's laws of electromagnetic induction.
State and explain Lenz's law. State and explain laws of electromagnetic force. Explain clearly how these forces are developed when two current carrying conductors have currents i same direction ii opposite direction. Develop an expression for force developed between two parallel current carrying conductors and hence define one ampere current.
Develop an expression for force between two long parallel current carrying conductors. Explain what you mean by dynamically induced emf. Develop an expression for the induced emf. Explain what you mean by statically induced emf. Hence develop an expression for induced emf for harmonically varying magnetic feld. Defne and explain what you mean by self inductance of coil. How do you compare the inductance of an air cored coil and an iron cored coil? Develop an expression for inductance of a coil in terms of its physical parameters.
Explain the variation of inductance of an iron coved coil as a function of current through the coil. Explain what you mean by mutual inductance between coils. Explain coefficient of magnetic coupling and develop an expression fr the same. Discuss its physical significance. Explain with neat diagram the process of Hysteresis in a magnetic material. How do you obtain magnetisation characteristic of a magnetic material from the hysteresis loops.
Develop an expresion for hysteresis loss in a magnetic material. Discuss various fctors which can be employed to reduce this loss. Explain with the help of neat diagram eddy current loss in a magnetic material. Discuss various methods to reduce thi s loss. Determine the voltage induced across the coil. Determinate flux in the coil. A particular coil has turns and an inductance of mH.
How many turns should be removed to get an inductor of 95 mH, keeping everything else unchanged? A 4-pole generator is running at rpm. The generated emf. Deter- mine the flux required at each pole. A turn coil is wound on a former having a rectangular cross-section of 5 cm x 2. The current in the coil is mA. Determine the inductance of the coil and the energy stored in the magnetic field.
Two coils of and turns respectively are wound on a common iron magnetic circuit of reluctance units. Determine the mutual inductance. Neglect leakage flux [Note: Two coils having 30 and turns respectively are wound side by side on a closed iron circuit of section cm2 and mean length ems.
With 4. Determine the mean value of the inductance between these current limits and the induced emf. A magnet has a residual magnetism in the air gap of 1. The pole face at the air gap is a square section 5 cm x 5 cm. How long time must a conductor be moved through the air gap to produce a generated emf of 10 m V. Assume that these is no fringing in the ai gap. Wadhwa C. Fitzerald A. Sen P. We have various waveforms viz. However, with L and C as the element, it is only sinusoidal waveform which satisfies the requirement of identical waveforms for source and response.
If the source is voltage source the current through the inductor is proportional to the integral of the voltage wave and for the capacitor, it is the derivative of the voltage wave.
It is a very useful and important property of sine waves that both it's derivative and integrals are also sinusoidal. Also, it is very easy to generate the sine waves using rotating machines and transmit and distribute electric energy using transformers.
We next illustrate that a uniform circular motion or simple harmonic motion is a sinusoidal variation. Sine wave is the basic waveform of a. Let us relate a uniform circular motion or rotation to a sinusoidal wave form.
Im and is coincident along x-axis Fig. We start counting time and rotate the phasor anti-clockwise and at various instant of time we take the projection of the rotating phasor along y-axis. After certain time say t the phasor rotates through an angle 8 and there is some finite projection along y-axis, mark this on the right hand side along t 9 axis as shown in Fig.
Mark it on the right hand side along t 8 axis as shown. A sinusoidal waveform having a displacement of 8 to the left is shown in Fig.
The average value of a cycle of a waveform is the area under the waveform divided by the length of one cycle. These are all sine waves In a since it is a full sine wave, the total area under the curve is zero as half of it is above ft axis and equal half is below ft-axis and hence the average area of the curve is zero and hence over the complete sine wave the average value is zero.
This is represented mathematically as. This is the efective value. We again take first, various sinusoidal waves as shown in Fig. As per the definition we have first to express the wave as per its instantaneous value as every instantaneous value is to be squared and then area under this curve is to be obtained. For Fig. Similarly the average value of a d.
If when F. In fact there are many other operating voltages e. So it is the standardization which has selected certain voltages as operating voltages.
Form factor does not have much of significance except when two wave frms one with higher form factor when applied to magnetic materials, produces higher hysteresis loss per loop as compared to one with lower form factor.
Crest factor has an indication of electric stresses developed in insulating material as the stresses are a function of the peak value of the wave.
The same aspect can be represented by two phasors rotating counter clockwise at the angular velocity u. Therefore, either the time waveform of the rotating phasors or the phasor diagram, can be used to describe the system. Since both the diagrams, the time diagram and the phasor diagram convey the same information, the phasor diagram being much more simpler, it is used for an explanation in circuit theory analysis.
A-C C 0Co 13 57 1. This can be represented mathematically in the following ways: Multiplication and division is very simple in polar coordinates. The term rL is known as inductive reactance as it is due to an inductance and we see that the two phasors v and i for an inductor are not identical.
If we compare expression for v and i we find that current i lags the voltage exactly by goo as shown in Fig. Vm I m drt - 0 2n I 2 This shows that the power consumed in a purely inductive circuit is zero. Here represents reactance of the capacitor. The sum of the voltage across R and ] should equal the supply voltage V. It is to be noted that if we divide the phasor diagram by I we will have another diagram as shown in Fig.
The product of V and 1 is known as apparent power. Hence power factor can be defined as P Real Power cos 8 - -. V Apparent Power The component of current along the voltage phasor I cos 8 gives real power or actual power whereas the component of current at right angle to the voltage phasor gives what is known as reactive power i. In fact reactive power is associated with the energy storing elements L and C. The analogy for this can he drawn between force applied to an object.
Find current and the power. Taking voltage as the reference we have. This shows that the current lags the voltage by an angle - 4.
Taking voltage as reference Impedance. A 1 5 V source is applied to a capacitive circuit that has an impedance of 0 -j 25 Q Determine the current and the power in the circuit. This is why, capacitor is also known as reactive power generator. The current can also be obtained using polar co-ordinates. Taking current as reference, the phasor diagram for the circuit is shown in Fig.
Example 1. A Naturally we have to connect some circuit element R, L or C so that the bulb gets.. The resistance of the lamp i s given by or or or. It is also given that Q of the coil is. Now current in the circuit is.. Hence impedance of the circuit is..
Also that impedance is. Hence, the phasor diagram in Fig. The same observation can also be made as follows: The units of admittance is mho or siemens and denoted by U or S respectively. Also from equation 1. The impedance of a parallel RC circuit is 0 -, The admittance of the circuit is. Determine the circuit i mpedance and the line current. Current Total current..
Since it is an inductive circuit, current I L will lag the voltage across it supply voltage here by angle 8. Determine the line current and the total impedance and admittance of the circuit shown in Fig. Whenever the natural frequency of oscillation of a system could be electrical, mechanical or a civil structure or a hydraulic coincides with the frequency of the driving force a voltage source in an electric circuit or a wind frce in a civil structure etc.
This phenomenon is known as resonance. This phenomenon may be useful under certain conditions and sometimes it may prove to be disastrous for the system. There are many engineering applications of resonance. A suspension bridge in Washington showed tendencies to oscillate up and down during construction and only a few months after construction, it began to build up oscillations under a moderate wind and within an hour the multi- billion dollar bridge was reduced to pieces.
This is a typical example of designing a bridge ignoring the possibility of phenomenon of resonance on the bridges. In case of a series compensated power system if the difference between the normal power frequency and the natural frequencies of oscillation of the system after compensation coincides with one of the natural torsional frequencies of the machine's shaft system Turbine, or combination of turbine and generator , torsional oscillation may be excited which may build up sufficiently to break the shaft of these machines.
A power transfrmer is normally operated at the knee-point of its B-H curve due to economic reasons. If some how the system voltage increases due to Ferranti effect or due to rejection of load, the transformer operates along the saturation region of the B-H curve where the inductance offered by the winding decreases and keeps on changing depending upon the dynamic behaviour of increase of system voltage. This is known as Ferro-resonance as it is due to the non-linear behaviour of the iron core of the transformer.
In such a situation it may be necessary to disconnect shunt capacitors very quickly to reduce the chances of occurrences of Ferro-resonance. In an electric circuit with inductance and capacitor in series, there is always a frequency at which the two reactances just cancel resulting in the minimum impedance resistive circuit characteristic of series resonance.
In electronic circuits many a times resonance condition is desired for maximum response for a given magnitude of excitation. In general there are two types of resonance in the electric circuits. We now study basic features and characteristic of these phenomena. However, it is to be noted that the phase relations between the voltage and current in the individual elements R, L and C are not same.
The net impedance is shown a positive quantity. The variation of current is also shown in Fig. The steepness of the response curve can be varied by varying the resistance of the circuit.
There are various applications of a series resonant circuit where the frequency is fixed and either L or C is varied to obtain the condition of resonance.
A typical example is that of tuning a radio receiver to a particular desired station that is operating at a fixed frequency. Here a circuit or L and C is adjusted to resonance at the operating fequency of the desired station. The capacitor C parallel plate is variable in most portable radio receivers and the inductance of the coil is usually varied in tuning of an automobile radio receiver. VL VL. Since the circuit at this frequency is capacitive the net voltage lags behind the circuit current by some angle.
Next we find expression for voltage drops across individual elements under resonance condition. Usually Q 1, hence it is also known as voltage gain, as the voltage across the inductor is much greater than the supply voltage.
If the capacitance is halved, the frequency of resonance would be. In electric circuit energy is stored in the fr! It can be proved that at any instant at a certain fequency the sum of energy stored by the inductor and the capacitor is A-C C 0Co 13 75 constant. Q is also a measure of the frequency selectivity of the circuit.
A circuit with high Q will have a very sharp current response curve as compared to one which has a low value of Q. To understand this let us consider Fig. Here we find that the current response is maximum at f0 and on either side of f0, the current decreases sharply. In order to obtain quantitative analysis of this reduction in current, we specify two fequencies f1 and f 2 at which the magnitude X L - Xe is equal to.
On subtracting. This means the resonance frequency is the geometric mean of the half-power frequen- Now on addition of. Bandwidth is thus given hy the ratio of the frequency of resonance to the quality factor and selectivity is defined as the ratio of resonant frequency to the bandwidth. The impedance of the circuit at resonance is Z. Frequency and Fig. Befre we move to parallel resonance, let us summarize the characteristics of series resonance circuit. At resonance power factor of the circuit is unity.
Therefore, supply voltage and current are in phase. The reactive component of the input impedance is zero and hence the circuit is resistive and hence current drawn hy the circuit is maximum.
The voltage across each reactive element is QV where Q is the quality factor of the element and Vis the supply voltage. The quality factor 7. For the circuit shown in Fig. A coil having a 5 ohm resistor is connected in sen: The circuit resonates at 1 00 Hz a Determine the i nductance o f the coil.
In series with these elements, resistances shown are their own resistances. In general there could he external resistance in series with these elements. A-C C 0Co 13 or or or or Fig. The input admittance is the sum of the admittances of the two branches. L L 1 1 L L 1 1 Now at resonance the imaginary part susceptance of admittance is zero. Therefore, Xco XLo r.
The equivalent input admittance is real and is given as: The resonance frequency for the circuit is given by the following expression when. It is seen that at resonance the circuit has maximum impedance and hence the current is minimum. At this frequency since the susceptance of the circuit is zero, the power factor of the circuit is unity. If we now consider that in the general circuit if.
L c Fig. This means there will be current in each parallel path but the net current from the supply is zero i. The supply current is 10 mA which supplies the losses of the parallel resonance circuit. The currents through the parallel branches gets magnified at resonance and is QI0. At resonance there will be exchange of energy between the inductor and the capacitor.
When the inductor is carrying maximum current, the voltage across capacitor is zero and hence the electrostatic energy is zero. R Therefore, and bandwidth.
Similarly using the equations 1. Summarising the characteristics of a parallel resonance we have: At resonance the input impedance is maximum or the input admittance is minimum and the circuit is purely resistive and hence power factor is unity. The current through the inductor and capacitor are equal i n magnitude but opposite in phase and the current is amplified by the factor Q, hence the current in each branch is QJ0 where 1 0 is current from the source.
Hence proved. R Bandwidth. In the problem of example 1. Determine the value of. Hence no value of. A coil has an inductance of 1. If the bandwidth required is 50 KHz what resistor should be connected across the coi l? H where His the magnetic field intensity and its units are Ampere turns per meter.
The units of mmf are ampere-turns when current is passed through N no. The direction of flux lines is given by cork-skrew rule or Fleming right-hand thumb rule. Similarly according to Fleming right hand thumb rule means if we hold our thumb in the direction of flow of current; here normal and into the paper the twist in the fingers gives the direction of fux lines on the surface of the paper.
However, in an electric circuit the voltage source is a part of the closed path, in the magnetic circuit the current carrying coil will surround or link the magnetic circuit. In tracing a magnetic circuit we will not be able to identify a pair of terminals at which the magnetomotive frce mmf is applied.
The analogous quantities are listed below. Electric Circuits e. No specific name is given to the dimension of reluctance except to refer it to as so many units of reluctance.
From the above expression for reluctance it is found that the impediment to the flow of flux which a magnetic circuit presents is directly proportional to the length and inversely proportional to the permeability and cross-sectional area.
The results are in consistence which the actual physical conditions. Series Magnetic Circuit If there are one or more than one magnetic paths connected in series there could even be an air gap in the path of the same flux same current in a series electric circuit , the net reluctance is the sum total of the individual magnetic paths.
R, Fig. Anal ogous paral l el electric circuits. Let us take a few examples to illustrate the application of what we have studied about the magnetic circuits. An iron ring of mean length 50 ems has an air gap of 1 mm and a winding of turns. If the relative permeability of air is when a current of 1 A fows through the coil, find the flux density. The reluctance of magnetic path R, Now reluctance of air gap R 2 4n?
A 95 Example 1. A cast steel electromagnet has an air gap of length 2 mm and an iron path of length 30 ems. Find the number of ampere turns necessary to produce a flux density of 0. Assume H - AT! A 1 Te central limbs has an air gap of 2 mm. The analogous electric circuit is as shown. R Total reluctance Hence Fig. A turn coil is wound on the central limb of the cast steel frame shown.
A total fux of 1. Find the current required. Dimensions are in ems. Reluctance for the air gap Reluctance of the central limb Fi g. SIeel Core Fig. If the secondary winding is open it draws current corresponding to exciting current which gives rise to magnetic flux in the steel core and meets the core loss requirements. Any magnetic fux that is outside the confines of the magnetic steel core is termed as leakage flux.
These components have been identified in the basic transformer in Fig. In an actual transformer with interleaved windings usually low voltage winding is placed near the core and then sufficient insulation is provided and high voltage is placed above the primary i. Since the leakage path is mainly through the air, the leakage flux and the voltage induced by it vary linearly with primary current. The effect on the primary circuit is simulated by assigning to the primary a leakage inductance which by definition is equal to leakage flux linkage leakage flux x No.
In the air gap the magnetic flux line bulge outwards somewhat as shown in Fig. The effect of the fringing is to increase the cross sectional area of air gap. The usual correction is to add the gap dimension to each dimension of air gap C. Suppose a time varying voltage source is connected across the primary of the transformer and at any given instant the voltage source has the polarity shown and current i t is in the direction shown by the arrow and is increasing with time entering the dotted terminal, this current induces a voltage in the secondary which is positive at the dotted terminal.
A ci rcui t for dot convention. The following experimental procedure can be used to establish dotted ends of the transformer windings. Mark a dot arbitrarily on one end of the primary winding and connect a d. Connect a MC voltmeter across the secondary winding.
The end of the secondary winding which goes positive momentarily on closing the switch on the primary side as measured by the voltmeter, is the terminal to be dotted. If there are more than two windings, similar procedure can be followed for identifying relationship between each pair of windings.
Suppose we are required to write loop equations for a mutually coupled circuit shown in Fig. Coupl ed circuit.
The voltage induced in primary will be determined by the direction of i 2 with respect to its dot in secondary loop. Since current is going away from the dot, the polarity of voltage on the primary side will be positive at the undotted terminal.
Therefore, for current i 1 it will be rise in voltage and the equation will be and for the secondary loop, similarly.
Equival ent of Fi g. R c Fig. Magneti cal l y coupl ed network. The equivalent with dots is given in Fig. Equi val ent of Fi g. Due to coil 3, since the current i3 is leaving the triangular dot, the polarity will be -ve at the corresponding dot in coil 1. Voltage induced i n coil 2 Due to coil 1, the current i 1 - i 2 enters the dot, therefre positive polarity at the dot of coil 2.
Voltage induced in coil 3 Due to coil 1 the current entering the dot is i 1 - i2 hence the positive polarity at the dotted terminal of coil 3. Due to coil 2 current i 2 - i3 leaves the dot square , the positive polarity will be at the undotted terminal.
Hence the equivalent circuit given as follows: The loop equations can be written as follows: Loop 1 Loop 2 Loop 3 R. When the two coils are to be interconnected it is important to know whether the mutual inductance Mis aiding or opposing. However, it is known that the effect on each of the two coils must be the same. Refer to Fig. If L 1 and L 2 are the self inductances of the two coils and M the mutual inductance, the total inductance say if is L A in case of Fig.
The coils at ri ght angl es. The co-efficient of coupling and the mutual inductance between the two coils shown in Fig. Although the fu of one coil appears to pass through the other coil, it does not actually link the turns of the other coil. Therefore, the linkages are zero and the value of Mis zero. It is only when one coil is rotated with respect to other that a linking occurs.
Determine the inductance between the terminals for a three coil system shown below: From the Fig. A sinusoidally varying alternating current of frequency 50 Hz has a maximum value of 20 A. I , Solution: Resolving the two currents along x and y-axes.
Hence the sum of the two voltages peak value Hence and phase Hence. A-C C 0Co 13 1 07 Example 1. An iron cored coil takes 5 A when connected to 40 V de supply and takes 5 A when connected to V ac supply and consumes watts.
Determine i the impedance ii the iron loss iii the inductance of the coil iv p. When coil is connected to de supply it ofers only resistance d. A non-inductive load takes 1 0 A at 1 00 V Determine the inductance of a reactor to be connected in series in order that the same current be supplied from V 50 Hz mains. Determine the phase angle between the V mains and the current neglect resistance of the reactance.
H Ans. An iron cored coil of resistance 5 Q takes 10 A when connected to V 50 Hz supply and the power dissipated is W The coil is assumed to be equivalent to a series combination of R and L, determine i the iron loss ii the inductance at the given value of current and the iii p.
W and power dissipated in the resistance part of the coil is. A-C C 0Co 13 Solution: Determine the instantaneous energy stored in the capacitor and inductor b Q of the circuit. Determine the current through all the branches of the given network. The coil has turns and the mean core length is mm when the air gap lengths are 1. The magnetic field intensity is AT!
Since the air gap is small flux fringing can be neglected. Hence the fux density is same in both the core the air gap.
The mmf required for the core is x 0. A toroid is composed of three ferromagnetic materials, nickel-iron alloy, silicon steel and cast steel each having mean arc lengths 0.
The toroid has a coil of 1 00 turns. Since magnetic field intensities in the three materials are known, the total mmf required i 10 x 0. Since Therefore B 0.
For the toroid of the previous example, determine the magnetic flux approximately produced if the mmf applied is 35 AT. Determine the i mmf ii magnetic flux iii reluctance and iv flux density for the following case: A steel ring 30 cm mean diameter and of circular section 2 ems in diameter has an air gap 1 mm long: Neglect magnetic leakage. The mmf required for air gap is 1 x 0. Therefore, from solution point of view, whichever is smaller of the two should be used for analysis of a given network.
However, it may be mentioned here that there could be situations when it could be mandatory to use only a particular kind of variables viz. Determine the effective value for the waveform shown Fi g. VJJl 1. The power in a 20 ohm load is 10 W. What is the peak to peak sinusoidal voltage across the load. A resistor and a capacitor in series are connected to a V, 60 lZ supply. The impedance of the circuit is 86 ohms and the power consumed is also 86 W. Determine the value of R and L. A ohm resistor is in series with a capacitor across V, 60 Hz.
The power is W. Determine C and 1. The resistance of a coil is Q and its inductance 0. Determine the current, the p. Determine the equivalent series impedance of the load. The series combination is placed across a V, 60 Hz supply.
Determine the current in rectangular and polar coordinates. Two appliances are operated in parallel from a V supply. One draws 12 A at a lagging p. Determine the total current and the power. Determine the branch currents, the line current and the power delivered to the circuit if it is connected across 1 1 5 V supply. If the total current supplied is 15 A, what is the power taken by each branch? Discuss various characteristics of a series RLC resonant circuit. Show that the voltage across the inductor or capacitor in a series RLC circuit is Q times the supply voltage at resonance.
What are half-power frequencies in a series RLC resonance circuit? Derive an expression for bandwidth of the circuit. Discuss various characteristics of a general parallel RLC circuit at resonance. Derive resonance frequency when a coil is loss free b capacitor is loss free c both are loss free. Draw the equivalent circuit when Q of the circuit is high.
Derive an expression for Q of a parallel resonance circuit from first principles. A given parallel circuit has certain bandwidth. If it is required to obtain a lower bandwidth what measure should be taken to achieve this? Explain what you mean by selectivity of a circuit and explain how it can be improved. A 70 ohm resistor, a 5 mH coil and a 15 pF capacitor are in series across a 1 10 volt supply. Determine the resonant fequency the Q of the circuit at this frequency, the voltage across the capacitor at resonance.
A variable inductor is connected in series with a resistor and a capacitor. The circuit is connected to a volt 50 Hz supply and the maximum current obtainable by varying inductance is 0.
The voltage across the capacitor then is V. Determine the circuit constants. A coil of 20 Q resistance and 0. Find the frequency of resonance and the efective impedance at resonance. A series circuit consists of a 1 15 Q resistor, a 0. If the resonant frequency of the circuit is Hz determine 1 and the bandwidth. Three coils are interconnected with winding sense shown by dots.
The self inductance of the coils are 2 H each and the mutual inductance 1 H each. Two coils are wound on the same core and have negligible resistance as shown in the Fig.
Determine the current in the ohm resistor and its phase angle with respect to the applied voltage of volt at a frequency of Hz when the coils act i in the same sense ii in opposite sense.
A cast steel ring has a circular cross sections 3 cm in diameter and a mean circumference of 80 cm. The ring is uniformly wound with a coil of turns. A ring of cast steel has an external diameter of 24 ems and a square section of 3 cm side. Inside and across the ring an ordinary steel bar 18 x 3 x 0. When the exciting current is 2. Determine relative permeability of iron. P iron magnetic circuit has a uniform cross sectional area of 5 cm2 and a length of 25 cm.
A coil of turns is wound uniformly over the magnetic circuit. When the current in the coil is 1. Define the following terms. Magnetic flux density, magnetic field intensity, mmf, reluctance. Discuss briefly analogies between magnetic and electric circuits. A two limb magnetic circuit has magnetic path of 30 ems and an air gap of 5 mm.
The area of cross section of magnetic path is 9 cm 2. The relative permeability of core is and the circuit has turns wound on one limb. The total flux in the gap of a permanent magnet is 0. The diameter of the circular pole faces are Determine the values of B and Hin the gap. Assume that these is no finging. A long wire of radius R carries de current I ampere uniformly distributed over the cross section.
A long horizontal wire carrying 20 A i s placed crosswise to the earth's magnetic field. At what point does the magmatic field of the wire just cancel terrestrial magnetism? The central limb has a cross section of mm 2 and each of the outer limb has a cross section of mm2.
Also while considering currents at a node, the currents entering the node will be taken as positive and those leaving would be taken as negative. The algebraic sum of all branch voltage around any closed loop of a network is zero at all instants of time. Alternatively, Kirchhoffs voltage law can be stated in terms of voltage drops and rises as follows: KL is a consequence of law of conservation of energy as voltage is energy or work per unit charge.
We know that electronic current flows form negative potential to positive potential, the conventional current flows from positive potential to negative potential. Kirchhofs current law states that the algebraic sum of all currents terminating at a node equals zero at any instant of time. The two basic laws by Kirchhofs can be applied to solve any network irrespective of its complexity. One typical application could be to find out equivalence between two networks or given a network in some configurations, how to find its equivalent so that it could be used more conveniently.
The objective is to replace it by a single equivalent resistance From Fig. We will try here fr inductors. Say there are two inductors connected in parallel. Suppose we are given star connected elements as shown in Fig. Between a and b. From equation 2.
Suppose we want to transform star into delta. The numerator is same for all the three branches and is equal to the product of impedance taken two at a time i. Similarly for Z B it will be za and for Z c it will be z b Fig.
To find Za open the delta vertex at A. The two impedances associated at vertex A are Z A and Zc. The equivalent star.
Here denominator is same for all the three legs of the equivalent star. Example 2. Let the currents i1 and i 2 be flowing in the two loops as shown. If they are coming out to be positive, the directions remain as shown in the figure, otherwise one with negative value, the direction of this current is reversed. Determine the current i n each branch of the network shown i n Fig. These equations can be written in matrix form as follows [ 0. The sign is positive if current i1 and i2 in the common branch flow in the same direction and is taken as negative if the two currents oppose each other.
The above set of these equations can again be written in matrix form as follows: On solution the approximate values of il' i2 and i 3 are found to be 40 A, 10 A and - 10 A. Here whenever a voltage source is given it should be converted into a current source before nodal equations using KCL are written. Taking node D as reference, we are left with three nodes A, B and C. Voltages of nodes A and C with respect to reference node are known to be 30V and 40 V respectively. Taking node E as refrence and since the potentials of nodes C and D with respect to E are known hence only two independent nodes are A and B where the voltages he V A and V B respectively.
Paul Scherz. The Art of Electronics. Paul Horowitz. Customers who bought this item also bought. Product details Series: Networks and Systems Hardcover: English ISBN Tell the Publisher! I'd like to read this book on Kindle Don't have a Kindle? Share your thoughts with other customers. Write a customer review. Top Reviews Most recent Top Reviews.
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