Any region such as that shown between the plates in Fig. The direction of the field is defined as that of the force acting on a positive charge placed in the field. Such a field may be represented in magnitude and direction by lines of electric force drawn between the charged surfaces. Whenever a p. Figure 6. Electric lines of force often called electric flux lines are continuous and start and finish on point charges; also, the lines cannot cross each other.
When a charged body is placed close to an uncharged body, an induced charge of opposite sign appears on the surface of the uncharged body. This is because lines of force from the charged body terminate on its surface. Hence the force between two charged spheres in air with their centres 16 mm apart and each carrying a charge of C1. They are connected to opposite terminals of a battery of voltage V volts.
There is therefore an electric field in the space between the plates. If the plates are close together, the electric lines of force will be straight and parallel and equally spaced, except near the edge where fringing will occur see Fig. Electric field strength is also called potential gradient. However, it should be remembered that they are only aids to the imagination. The force of attraction or repulsion between two electrically charged bodies is proportional to Figure 6.
Thus the presence of the field indicates the presence of equal positive and negative electric charges on the two plates of Fig. The property of this pair of plates which determines how much charge corresponds to a given p. For example, there is capacitance between the conductors of overhead transmission lines and also between the wires of a telephone cable. In these examples the capacitance is undesirable but has to be accepted, minimized or compensated for. There are other situations where capacitance is a desirable property.
Devices specially constructed to possess capacitance are called capacitors or condensers, as they used to be called. In its simplest form a capacitor consists of two plates which are separated by an insulating material known as a dielectric. A capacitor has the ability to store a quantity of static electricity. The symbols for a fixed capacitor and a variable capacitor used in electrical circuit diagrams are shown in Fig. Calculate how long the capacitor can provide an average discharge current of 2 mA. Determine the capacitance of the capacitor.
Calculate the final p. Find the capacitance of the capacitor. Thus electric flux is measured in coulombs, and for a charge of Q coulombs, the flux D Q coulombs. Compared with conductors, dielectric materials have very high resistivities. They are therefore used to separate conductors at different potentials, such as capacitor plates or electric power lines.
Two parallel rectangular plates measuring 20 cm by 40 cm carry an electric charge of 0. Calculate the electric flux density. If the plates are spaced 5 mm apart and the voltage between them is 0. Find the voltage gradient between the plates. Two parallel plates having a p. What is the electric field strength? Find also the electric flux density when the dielectric between the plates is a air, and b polythene of relative permittivity 2.
What is the electric field strength across the dielectric at this voltage? If the effective area of each plate is 5 cm2 find the electric flux density of the electric field. If the plates are spaced 10 mm apart and the voltage between them is 0. Determine the electric field strength. Calculate the capacitance of the capacitor in picofarads. Ten plates are shown, forming nine capacitors with a capacitance nine times that of one pair of plates. Thus capacitance e0 er A.
A waxed paper capacitor has two parallel plates, each of effective area cm2. If the capacitance of the capacitor is pF determine the effective thickness of the paper if its relative permittivity is 2. A parallel plate capacitor has nineteen interleaved plates each 75 mm by 75 mm separated by mica sheets 0. Assuming the relative permittivity of the mica is 5, calculate the capacitance of the capacitor.
Calculate the capacitance in picofarads. If the capacitance is pF determine the effective thickness of the paper if its relative permittivity is 2 [0. If the capacitance of the capacitor is pF determine the thickness of the mica sheet. What would be the capacitance of the capacitor if the plate area is doubled and the plate spacing is halved? It has 19 plates, each 50 mm by 30 mm separated by a dielectric of thickness 0. Determine the relative permittivity of the dielectric. If the capacitor has twenty-five plates separated by a dielectric of thickness 0.
Find a the thickness of the polythene needed, and b the area of a plate. The capacitors each store a charge and these are shown as Q1 , Q2 and Q3 respectively. Note that this formula is similar to that used for resistors connected in series. Let the p. Hence when capacitors are connected in series the charge on each is the same. Note that this formula is similar to that used for resistors connected in parallel. Calculate a the equivalent circuit capacitance, b the charge on each capacitor, and c the p.
Determine a the equivalent circuit capacitance, b the total charge and c the charge on each capacitor. The reason for this can be seen from the above problem where the lowest valued capacitor i. Let this charge be Q coulombs. Then i. The circuit is now as shown in Fig. Determine the equivalent capacitance in each case. Find the total equivalent circuit capacitance.
What capacitance must be added in series to obtain a capacitance of 1. Determine the values of P, Q and R [4. Determine a the equivalent capacitance, b the total charge and c the charge on each capacitor. When the circuit is connected across a V d. Calculate a the equivalent capacitance, b the The maximum amount of field strength that a dielectric can withstand is called the dielectric strength of the material. A capacitor is to be constructed so that its capacitance is 0. Find a the thickness of the mica needed, and b the area of a plate assuming a two-plate construction. Find the p. A capacitor is charged with 10 mC.
If the energy stored is 1. Find a the capacitance and b the energy stored [ a 0. As the moving plates are rotated through half a revolution, the meshing, and therefore the capacitance, varies from a minimum to a maximum value. Variable air capacitors are used in radio and electronic circuits where very low losses are required, or where a variable capacitance is needed. The maximum value of such capacitors is between pF and pF.
If the energy stored is 0. Mica capacitors. A typical older type construction is shown in Fig. Calculate a the energy stored, b the electric flux density and c the potential gradient [ a 1. The main types include: variable air, mica, paper, ceramic, plastic, titanium oxide and electrolytic.
Variable air capacitors. These usually consist of two sets of metal plates such as aluminium , one fixed, the other variable. The set of moving Figure 6. Mica is easily obtained in thin sheets and is a good insulator. However, mica is expensive and is not used in capacitors above about 0.
A modified form of mica capacitor is the silvered mica type. The mica is coated on both sides with a thin layer of silver which forms the plates. Capacitance is stable and less likely to change with age. Such capacitors have a constant capacitance with change of temperature, a high working voltage rating and a long service life and are used in high frequency circuits with fixed values of capacitance up to about pF. Paper capacitors. A typical paper capacitor is shown in Fig. The whole is usually impregnated with oil or wax to exclude moisture, and then placed in a plastic or aluminium container for protection.
Disadvantages of paper capacitors include variation in capacitance with temperature change and a shorter service life than most other types of capacitor. Ceramic capacitors. These are made in various forms, each type of construction depending on the value of capacitance required. For high values, a tube of ceramic material is used as shown in the cross section of Fig. For smaller values the cup construction is used as shown in Fig. Certain ceramic materials have a very high permittivity and this enables capacitors of high capacitance to be made which are of small physical size with a high working voltage rating.
Ceramic capacitors are available in the range 1 pF to 0.
- Maximum Entropy and Bayesian Methods: Paris, France, 1992.
- A Feral Chirstmas.
- Electrical and Electronic Principles and Technology - PDF Free Download.
- Tag: Electrical Engineering Principles.
- Electrical Principles, 4th Edition by Peter Phillips | | Booktopia!
- Ultimate IQ tests : 1000 practice test questions to boost your brainpower.
Plastic capacitors. Some plastic materials such as polystyrene and Teflon can be used as dielectrics. Construction is similar to the paper capacitor but using a plastic film instead of paper. Plastic capacitors operate well under conditions of high temperature, provide a precise value of Figure 6. Titanium oxide capacitors have a very high capacitance with a small physical size when used at a low temperature. Construction is similar to the paper capacitor with aluminium foil used for the plates and with a thick absorbent material, such as paper, impregnated with an electrolyte ammonium borate , separating the plates.
The finished capacitor is usually assembled in an aluminium container and hermetically sealed. Its operation depends on the formation of a thin aluminium oxide layer on the positive plate by electrolytic action when a suitable direct potential is maintained between the plates. This oxide layer is very thin and forms the dielectric. The absorbent paper between the plates is a conductor and does not act as a dielectric. Such capacitors must always be used on d. Electrolytic capacitors are manufactured with working voltage from 6 V to V, although accuracy is generally not very high.
These capacitors possess a much larger capacitance than other types of capacitors of similar dimensions due to the oxide film being only a few microns thick. The fact that they can be used only on d. Thus precautions must be taken to ensure that the capacitor is automatically discharged after the supply is switched off. This is done by connecting a high value resistor across the capacitor terminals. The equivalent capacitance is. What is the capacitance of the capacitor? The capacitance of a capacitor a is proportional to the cross-sectional area of the plates b is proportional to the distance between the plates c depends on the number of plates 67 d is proportional to the relative permittivity of the dielectric 8 Which of the following statement is false?
A permanent magnet will position itself in a north and south direction when freely suspended. The north-seeking end of the magnet is called the north pole, N, and the south-seeking end the south pole, S. The area around a magnet is called the magnetic field and it is in this area that the effects of the magnetic force produced by the magnet can be detected.
A magnetic field cannot be seen, felt, smelt or heard and therefore is difficult to represent. Michael Faraday suggested that the magnetic field could be represented pictorially, by imagining the field to consist of lines of magnetic flux, which enables investigation of the distribution and density of the field to be carried out. The distribution of a magnetic field can be investigated by using some iron filings. A bar magnet is placed on a flat surface covered by, say, cardboard, upon which is sprinkled some iron filings.
If a number of magnets of different strength are used, it is found that the stronger the field the closer are the lines of magnetic flux and vice versa. Thus a magnetic field has the property of exerting a force, demonstrated in this case by causing the iron filings to move into the pattern shown. The strength of the magnetic field decreases as we move away from the magnet. It should be realized, of course, that the magnetic field is three dimensional in its effect, and not acting in one plane as appears to be the case in this experiment.
Figure 7. The direction of a magnetic field at any point is taken as that in which the north-seeking pole of a compass needle points when suspended in the field. The direction of a line of flux is from the north pole to the south pole on the outside of the magnet and is then assumed to continue through the magnet back to the point at which it emerged at the north pole.
Thus such lines of flux always form complete closed loops or paths, they never intersect and always have a definite direction. The laws of magnetic attraction and repulsion can be demonstrated by using two bar magnets. Lines of flux are imagined to contract and the magnets try to pull together.
Atlas of Human Anatomy by Netter
The magnetic field is strongest in between the two magnets, shown by the lines of flux being close together. The unit of magnetic flux is the weber, Wb Magnetic flux density is the amount of flux passing through a defined area that is perpendicular to the direction of the flux: Magnetic flux density D magnetic flux area The symbol for magnetic flux density is B.
A magnetic pole face has a rectangular section having dimensions mm by mm. The maximum working flux density of a lifting electromagnet is 1. If the total magnetic flux produced is mWb, determine the radius of the pole face. Flux density B D 1. Thus m. If the coil is uniformly wound around the circuit and has turns, find the current in the coil. If the total magnetic flux produced is mWb determine the radius of the pole face. If the coil has turns find the current in the coil. From its definition, r for a vacuum is 1. For nonmagnetic materials this is a straight line.
Typical curves for four magnetic materials are shown in Fig. The approximate range of values of relative permeability r for some common magnetic materials are: Cast iron Mild steel Silicon iron Cast steel Mumetal Stalloy r r r r r r D — D — D — D — D — D — Problem 4. A flux density of 1. Find the relative permeability of the steel under these conditions. Determine the magnetic field strength and the m.
- Refine your editions:.
- Completed Text Books.
- Bestselling Series.
- Dehydration of foods.
- Advanced Carbon Materials and Technology.
- The Warrior King (The Seer King Trilogy, Book 3);
- Mira Corpora;
A coil of turns is wound uniformly on a ring of non-magnetic material. The ring has a mean circumference of 40 cm and a uniform cross-sectional area of 4 cm2. If the current in the coil is 5 A, calculate a the magnetic field strength, b the flux density and c the total magnetic flux in the ring. A uniform ring of cast iron has a cross-sectional area of 10 cm2 and a mean circumference of 20 cm. Determine the m. The magnetisation curve for cast iron is shown on page Such a solution is shown below in Table 1. Problem 7. An iron ring of mean diameter 10 cm is uniformly wound with turns of wire.
When a current of 0. Find a the magnetising force and b the relative permeability of the iron under these conditions. From the magnetisation curve for cast iron, shown on page 71, derive the curve of r against H. The curve demonstrates the change that occurs in the relative permeability as the magnetising force increases. Find the current required in a coil of turns wound on the ring to produce a flux of 0.
Use the magnetisation curve for cast steel shown on page 71 [0. If the flux required in the air-gap is 0. Find the flux density now in the solenoid. Use the B—H curve for mild steel shown on page 71 b If a coil of turns is wound uniformly around the ring in Part a what current would be required to produce the flux? Ferromagnetic materials have a low reluctance and can be used as magnetic screens to prevent magnetic fields affecting materials within the screen. Determine the reluctance of a piece of mumetal of length mm and cross-sectional area mm2 when the relative permeability is 4 Find also the absolute permeability of the mumetal.
A mild steel ring has a radius of 50 mm and a cross-sectional area of mm2. A current of 0. If the relative permeability at this value of current is find a the reluctance of the mild steel and b the number of turns on the coil. Calculate a the reluctance and b the absolute permeability of the steel. If the relative permeability of the steel at this value of current is find a the reluctance of the material and b the number of turns of the coil. A closed magnetic circuit of cast steel contains a 6 cm long path of cross-sectional area 1 cm2 and a 2 cm path of cross-sectional area 0.
A coil of turns is wound around the 6 cm length of the circuit and a current of 0. Determine the flux density in the 2 cm path, if the relative permeability of the cast steel is For air, HD D 4. Total m. A tabular method could have been used as shown at the bottom of the page. The dimensions are: Problem A silicon iron ring of cross-sectional area 5 cm2 has a radial air gap of 2 mm cut into it. If the mean length of the silicon iron path is 40 cm calculate the magnetomotive force to produce a flux of 0.
The magnetisation curve for silicon is shown on page The total m. Determine also the total circuit reluctance. A section through a magnetic circuit of uniform cross-sectional area 2 cm2 is shown in Fig. The cast steel core has a mean length of 25 cm. The air gap is 1 mm wide and the coil has turns. The B—H curve for cast steel is shown on page Determine the current in the coil to produce a flux density of 0.
Determine the mmf necessary to establish a flux of 0. Use the B—H curve for cast steel shown on page When each of the air gaps are 1. Use the B—H curves shown on page A coil of 50 turns is wound around the 40 mm length of the circuit and a current of 0. Find the flux density in the 15 mm length path if the relative permeability of the silicon iron at this value of magnetising force is 3 The silicon iron magnetic circuit has a uniform crosssectional area of 3 cm2 and its magnetisation curve is as shown on page Use a tabular approach to determine the total m.
Find also the total reluctance of the circuit. Use the magnetisation curves shown on page The resulting relationship between B and H is shown by the curve Oab in Fig. The material is said to be saturated. Thus by is the saturation flux density. The area of a hysteresis loop varies with the type of material.
The area, and thus the energy loss, is much greater for hard materials than for soft materials. When H is reduced to zero, flux remains in the iron. This remanent flux density or remanence is shown as Oc in Fig. When H is increased in the opposite direction, the flux density decreases until, at a value shown as Od, the flux density has been reduced to zero. The magnetic field strength Od required to remove the residual magnetism, i. Further increase of H in the reverse direction causes the flux density to increase in the reverse direction until saturation is reached, as shown by curve de.
If H is varied backwards from Ox to Oy, the flux density follows the curve efgb, similar to curve bcde. It is seen from Fig. This effect is called hysteresis. Hysteresis loss A disturbance in the alignment of the domains i. This energy appears as heat in the specimen and is called the hysteresis loss The energy loss associated with hysteresis is proportional to the area of the hysteresis loop.
Thus a hysteresis loop with a large area as with hard steel is often unsuitable since the energy loss would be considerable. Silicon steel has a narrow hysteresis loop, and thus small hysteresis loss, and is suitable for transformer cores and rotating machine armatures. Mark the direction of the field.
Exercise 36 Multi-choice questions on magnetic circuits Answers on page 1 The unit of magnetic flux density is the: a weber b weber per metre c ampere per metre d tesla 2 The total flux in the core of an electrical machine is 20 mWb and its flux density is 1 T.
The cross-sectional area of the core is: a 0. The ring has a mean circumference of 1 m and a uniform crosssectional area of 10 cm2. The current in the coil is 1 A. The flux produced is: a 1 Wb b Wb c 1 mWb d From the following list, match the magnetic quantities with their equivalent electrical quantities. With this arrangement, the magnets will: a attract each other b have no effect on each other c repel each other d lose their magnetism TLFeBOOK Assignment 2 This assignment covers the material contained in Chapters 5 to 7.
If a 10 V supply voltage is connected across the arrangement determine the current flowing through and the p. Assume electrical energy costs 7. A supply p. The ring has a uniform cross-sectional area of mm2 and a mean circumference of mm. If the current in the coil is 4 A, determine a the magnetic field strength, b the flux density, and c the total magnetic flux in the ring.
The voltage between the plates is V. Calculate a the electric flux density b the electric field strength, and c the capacitance of the capacitor, 7 A mild steel ring of cross-sectional area 4 cm2 has a radial air-gap of 3 mm cut into it. If the mean length of the mild steel path is mm, calculate the magnetomotive force to produce a flux of 0. Use the B—H curve on page 71 8 Figure A2. Let a piece of wire be arranged to pass vertically through a horizontal sheet of cardboard on which is placed some iron filings, as shown in Fig.
If a current is now passed through the wire, then the iron filings will form a definite circular field pattern with the wire at the centre, when the cardboard is gently tapped. By placing a compass in different positions the lines of flux are seen to have a definite direction as shown in Fig. Figure 8. The effect on both the iron filings and the compass needle disappears when the current is switched off. The magnetic field is thus produced by the electric current. The magnetic flux produced has the same properties as the flux produced by a permanent magnet.
If the current is increased the strength of the field increases and, as for the permanent magnet, the field strength decreases as we move away from the current-carrying conductor. If the whole length of the conductor is similarly investigated it is found that the magnetic field round a straight conductor is in the form of concentric cylinders as shown in Fig. This may be thought of as the feathered end of the shaft of an arrow. See Fig. This may be thought of as the point of an arrow. For example, with current flowing away from the viewer Fig. Hence the direction of the magnetic field is clockwise.
A magnetic field set up by a long coil, or solenoid, is shown in Fig. If the solenoid is wound on an iron bar, as shown in Fig. The direction of the magnetic field produced by the current I in the solenoid may be found by either of two methods, i. The direction of the magnetic field inside the solenoid is from south to north. Thus in Figures 4 a and b the north pole is to the right. The magnetic field associated with the solenoid in Fig. Thus the north pole is at the bottom and the south pole at the top. An electromagnet, based on the solenoid, provides the basis of many items of electrical equipment, examples of which include electric bells, relays, lifting magnets and telephone receivers.
Sketch the magnetic field pattern associated with the current carrying coil and determine the polarity of the field. A typical single stroke bell circuit is shown in Fig. Since the ironcored coil is energised the soft iron armature is attracted to the electromagnet.
The armature also carries a striker which hits the gong. When the circuit is broken the coil becomes demagnetised and the spring steel strip pulls the armature back to its original position. The striker will only operate when the push button is operated. A typical simple relay is shown in Fig. When the coil is energised the hinged soft iron armature is attracted to the electromagnet and pushes against two fixed contacts so that they are connected together, thus closing some other electrical circuit.
The load, Q, which must be of magnetic material is lifted when the coils are energised, the magnetic flux paths, M, being shown by the broken lines. A typical telephone receiver is shown in Fig. A thin, flexible diaphragm of magnetic material is held in position near to the magnetic poles but not touching them.
Variation in current from the transmitter varies the magnetic field and the diaphragm consequently vibrates. The vibration produces sound variations corresponding to those transmitted. A typical robust lifting magnet, capable of exerting large attractive forces, is shown in the elevation and plan view of Fig.
Over the face of the electromagnet is placed Figure 8. The force on the current-carrying conductor in a magnetic field depends upon: a the flux density of the field, B teslas b the strength of the current, I amperes, c the length of the conductor perpendicular to the magnetic field, l metres, and d the directions of the field and the current. When the magnetic field, the current and the conductor are mutually at right angles then: Figure 8.
Loudspeaker A simple application of the above force is the moving coil loudspeaker. The loudspeaker is used to convert electrical signals into sound waves. A moving coil, called the voice or speech coil, is suspended from the end of a paper or plastic cone so that it lies in the gap. When an electric current flows through the coil it produces a force which tends to move the cone backwards and forwards according to the direction of the current. The cone acts as a piston, transferring this force to the air, and producing the required sound waves.
This is the basic principle of operation of the electric motor see Section 8. A conductor carries a current of 20 A and is at right-angles to a magnetic field having a flux density of 0. If the length of the conductor in the field is 30 cm, calculate the force acting on the conductor. A conductor mm long carries a current of 10 A and is at right-angles to a magnetic field lying between two circular pole faces each of radius 60 mm. If the total flux between the pole faces is 0.
ThuMb - Motion l D mm D 0. Determine the current required in a mm length of conductor of an electric motor, when the conductor is situated at right-angles to a magnetic field of flux density 1. If the conductor is vertical, the current flowing downwards and the direction of the magnetic field is from left to right, what is the direction of the force? Force D 1. The lines of flux will reinforce i. Hence the force on the conductor will be from back to front i. With reference to Fig. The current is flowing towards the viewer, and using the screw rule, the direction of the field is anticlockwise.
If the length of the conductor in the field is mm calculate the force acting on the conductor. A coil is wound on a rectangular former of width 24 mm and length 30 mm. The former is pivoted about an axis passing through the middle of the two shorter sides and is placed in a uniform magnetic field of flux density 0. If the coil carries a current of 50 mA, determine the force on each coil side a for a single-turn coil, b for a coil wound with turns.
Thus the total force produced by the current is times that for 2 Calculate the current required in a mm length of conductor of a d. Calculate the strength of the magnetic field if a current of 15 A in the conductor produces a force on it of 3. The poles have a circular crosssection. If the force exerted on the conductor is 80 N and the total flux between the pole faces is 1. If the coil carries a current of mA, determine the force exerted on each coil side, a for a single-turn coil, b for a coil wound with turns. When current flows in the coil a magnetic field is set up around the coil which interacts with the magnetic field produced by the magnets.
This causes a torque and the coil rotates anticlockwise. If the current is not Figure 8. The current direction is reversed every time the coil swings through the vertical position and thus the coil rotates anti-clockwise for as long as the current flows. This is the principle of operation of a d. When a conductor carrying current is placed in a magnetic field, a force F is exerted on the conductor, given by F D BIl. If the flux density B is made constant by using permanent magnets and the conductor is a fixed length say, a coil then the force will depend only on the current flowing in the conductor.
In a moving-coil instrument a coil is placed centrally in the gap between shaped pole pieces as shown by the front elevation in Fig. The air-gap is kept as small as possible, although for clarity it is shown exaggerated in Fig. Current is led into and out of the coil by two phosphor bronze spiral hairsprings which are wound in opposite directions to minimize the effect of temperature change and to limit the coil swing i. Current flowing in the coil produces forces as shown in Fig. The two forces, FA and FB , produce a torque which will move the coil in a clockwise direction, i.
Since force is proportional to current the scale is linear. When the aluminium frame, on which the coil is wound, is rotated between the poles of the magnet, small currents called eddy currents are induced into the frame, and this provides automatically the necessary damping of the system due to the reluctance of the former to move within the magnetic field.
An electron in a television tube has a charge of 1. Determine the force exerted on the electron in the field. Show the direction of the field. Name them. For the direction of current flow shown in the coil determine the direction that the pointer will move. Briefly explain the principle of operation of such an instrument. If the current flows into the coil at C, the coil will: a commence to rotate anti-clockwise b commence to rotate clockwise c remain in the vertical position d experience a force towards the north pole Exercise 40 Multi-choice questions on electromagnetism Answers on page 1 A conductor carries a current of 10 A at right-angles to a magnetic field having a flux density of mT.
If the length of the conductor in the field is 20 cm, the force on the conductor is: a kN b 1 kN c N d 1 N 2 If a conductor is horizontal, the current flowing from left to right and the direction of the surrounding magnetic field is from above to below, the force exerted on the conductor is: a from left to right b from below to above c away from the viewer d towards the viewer 3 For the current-carrying conductor lying in the magnetic field shown in Fig. The electron has a charge of: a 1. If the conductor forms part of a closed circuit then the e.
Hence an e. Figure 9. If the first finger points in the direction of the magnetic field and the thumb points in the direction of motion of the conductor relative to the magnetic field, then the second finger will point in the direction of the induced e. As the magnet is moved towards the coil, the magnetic flux of the magnet moves across, or cuts, the coil. It is the relative movement of the magnetic flux and the coil that causes an e.
This effect is known as electromagnetic induction. The laws of electromagnetic induction stated in section 9. A generator converts mechanical energy into electrical energy. The action of a simple a. The induced e. E set up between the ends of the conductor shown in Fig. At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0. Assume the conductor, the field and the direction of motion are mutually perpendicular. Induced e. When a conductor moves in a magnetic field it will have an e. The wing span of a metal aeroplane is 36 m.
The diagrams shown in Fig. Determine i the direction in which the conductor has to be moved in Fig. Hence the force on the conductor is to the right. Thus the conductor will have to be moved to the left. S, or right to left; ThuMb - Motion, i. GeneRator rule. Determine the current flowing in the conductor when a its ends are open-circuited, b its ends are connected to a load of 15 ohms resistance.
Given that the e. If the conductor forms part of a closed circuit of total resistance 5 ohms, calculate the force on the conductor. Assuming the back axle of the car is 1. If the flux on the pole face is 60 mWb, find the magnitude of the induced e. The unit of inductance is the henry, H. A circuit has an inductance of one henry when an e. Find the average e. When the e. Calculate the e.
Determine the inductance of the coil. Find the time, in milliseconds, in which the flux makes the change. What rate of change of flux is required for this to happen? Find the magnitude of the average e. An average e. Calculate the time taken for the current to reverse. The basic form of an inductor is simply a coil of wire. Factors which affect the inductance of an inductor include: i the number of turns of wire — the more turns the higher the inductance ii the cross-sectional area of the coil of wire — the greater the cross-sectional area the higher the inductance iii the presence of a magnetic core — when the coil is wound on an iron core the same current sets up a more concentrated magnetic field and the inductance is increased iv the way the turns are arranged — a short thick coil of wire has a higher inductance than a long thin one.
An 8 H inductor has a current of 3 A flowing through it. How much energy is stored in the magnetic field of the inductor? Coil of wire Energy stored, b Figure 9. Inductance is often undesirable in a circuit. To reduce inductance to a minimum the wire may be bent back on itself, as shown in Fig. The wire may be coiled around an insulator, as shown, without increasing the inductance. Standard resistors may be non-inductively wound in this manner. Find the energy stored in the magnetic field of the inductor. Calculate the inductance of the coil.
Then, from section 9. Calculate the coil inductance when a current of 4 A in a coil of turns produces a flux of 5 mWb linking with the coil. A turn coil of inductance 3 H carries a current of 2 A. Calculate the flux linking the coil and the e. A flux of 25 mWb links with a turn coil when a current of 3 A passes through the coil. Calculate a the inductance of the coil, b the energy stored in the magnetic field, and c the average e. When a current of 1. If the coil inductance is 0. Thus I 0.
Calculate the coil inductance and the e. What current must flow to set up a flux of 20 mWb? How many turns would be needed to produce a 0. The current of 5 A is then reversed in Thus mutual inductance, M D 1. The mutual inductance between two coils is 18 mH. Calculate the steady rate of change of current in one coil to induce an e. Two coils have a mutual inductance of 0. If the current in one coil is changed from 10 A to 4 A in 10 ms, calculate a the average induced e. The phenomenon of mutual inductance is used in transformers see chapter 21, page Problem Find the magnitude of the e.
Calculate the magnitude of the e. If the current in one coil changes from 15 A to 6 A in 12 ms, calculate a the average e. If a current of 6 A in one coil is reversed in 0. E induced in a moving conductor may be calculated using the formula E D Blv. Name the quantities represented and their units 7 What is self-inductance? E is given by E D. State its symbol 13 The mutual inductance between two coils is M. The magnet is now withdrawn along the same path at 0. The deflection of the galvanometer is in the: a same direction as previously, with the magnitude of the deflection doubled b opposite direction as previously, with the magnitude of the deflection halved c same direction as previously, with the magnitude of the deflection halved d opposite direction as previously, with the magnitude of the deflection doubled 3 When a magnetic flux of 10 Wb links with a circuit of 20 turns in 2 s, the induced e.
The effective length of the conductor in the magnetic field is: a 20 cm b 5 m c 20 m d 50 m 6 Which of the following is false? What is the effect produced on the coil after a short time? In order to detect electrical quantities such as current, voltage, resistance or power, it is necessary to transform an electrical quantity or condition into a visible indication. A mechanical force is produced by the current or voltage which causes the pointer to deflect from its zero position. The controlling force acts in opposition to the deflecting force and ensures that the deflection shown on the meter is always the same for a given measured quantity.
It also prevents the pointer always going to the maximum deflection. There are two main types of controlling device — spring control and gravity control. The damping force ensures that the pointer comes to rest in its final position quickly and without undue oscillation. There are three main types of damping used — eddycurrent damping, air-friction damping and fluidfriction damping. There are basically two types of scale — linear and non-linear.
A linear scale is shown in Fig. The voltmeter shown has a range 0— V, i. A nonlinear scale is shown in Fig. The ammeter shown has a f. Figure When current passes through the solenoid, the two pieces of iron are magnetized in the same direction and therefore repel each other. The pointer thus moves across the scale. The force moving the pointer is, in each type, proportional to I2 and because of this the direction of current does not matter.
The moving-iron instrument can be used on d. When A moving-coil instrument, which measures only d. The average value of the full wave rectified current is 0. However, a meter being used to measure a. For the principle of operation of a moving-coil instrument, see Chapter 8, page Rectifier instruments have scales calibrated in r. There is no difference between the basic instrument used to measure current and voltage since both use a milliammeter as their basic part. This is a sensitive instrument which gives f.
When an ammeter is required to measure currents of larger magnitude, a proportion of the current is diverted through a lowvalue resistance connected in parallel with the meter. Such a diverting resistor is called a shunt. From Fig. Problem 2. A moving-coil instrument having a resistance of 10 , gives a f. Calculate the value of the multiplier to be connected in series with the instrument so that it can be used as a voltmeter for measuring p.
A moving-coil instrument gives a f. Calculate the value of the shunt to be connected in parallel with the meter to enable it to be used as an ammeter for measuring currents up to 50 A The circuit diagram is shown in Fig. Now try the following exercise Exercise 48 Further problems on shunts and multipliers 1 A moving-coil instrument gives f. Neglecting the resistance of the instrument, calculate the approximate value of series resistance needed to enable the instrument to measure up to a 20 V b V c V [ a 2 k b 10 k c 25 k ] 2 A meter of resistance 50 has a f.
Determine the value of shunt resistance required in order that f. Calculate the values of resistance required to enable the instrument to be used a as a 0—10 A ammeter, and b as a 0— V voltmeter. State the mode of resistance connection in each case. Calculate the value of resistance that converts the movement into a an ammeter with a maximum deflection of 50 A b a voltmeter with a range 0— V [ a The digital voltmeter DVM is one which provides a digital display of the voltage being measured. Advantages of a DVM over analogue instruments include higher accuracy and resolution, no observational or parallex errors see section A digital multimeter is a DVM with additional circuitry which makes it capable of measuring a.
Instruments for a. Some instruments, such as the moving-iron and electro-dynamic instruments, give a true r. With other instruments the indication is either scaled up from the mean value such as with the rectified moving-coil instrument or scaled down from the peak value. Sometimes quantities to be measured have complex waveforms see section Such waveform errors can be largely eliminated by using electronic instruments. A simple ohmmeter circuit is shown in Fig. Unlike the ammeter or voltmeter, the ohmmeter circuit does not receive the energy necessary for its operation from the circuit under test.
In the ohmmeter this energy is supplied by a self-contained source of voltage, such as a battery. Initially, terminals XX are short-circuited Thus f. The milliammeter can thus be calibrated directly in ohms. When calibrated, an unknown resistance is placed between terminals XX and its value determined from the position of the pointer on the scale.
An ohmmeter designed for measuring low values of resistance is called a continuity tester. An ohmmeter designed for measuring high values of resistance i. If a battery is incorporated then resistance can also be measured. Such instruments are called multimeters or universal instruments or multirange instruments.
A particular range may be selected either by the use of separate terminals or by a selector switch. Only one measurement can be performed at a time. Often such instruments can be used in a. The instrument has two coils: i a current coil, which is connected in series with the load, like an ammeter, and ii a voltage coil, which is connected in parallel with the load, like a voltmeter. A voltmeter should have as high a resistance as possible — ideally infinite. Calculate the power dissipated by the voltmeter and by resistor R in Fig. Thus the power dissipated in the voltmeter is insignificant in comparison with the power dissipated in the load.
In this case the higher load resistance reduced the power dissipated such that the voltmeter is using as much power as the load. An ammeter has a f. The ammeter is used to measure the current in a load of resistance when the supply voltage is 10 V. Calculate a the ammeter reading expected neglecting its resistance , b the actual current in the circuit, c the power dissipated in the ammeter, and d the power dissipated in the load. A voltmeter having a f. When the voltmeter is connected across the 40 k resistor the circuit is as shown in Fig.
Thus the ammeter itself has caused the circuit conditions to change from 20 mA to The error is reduced by using a voltmeter with a higher sensitivity. Determine the power dissipated in the load. What is the voltage across the 1. Determine the voltage indicated. Calculate: a the approximate value of current neglecting the ammeter resistance , b the actual current in the circuit, c the power dissipated in the ammeter, d the power dissipated in the 1 k resistor. Determine the wattmeter reading assuming the current in the load is still 15 A.
For examining periodic waveforms the electron beam is deflected horizontally i. The signal to be examined is applied to the vertical deflection system Y direction usually after amplification. Oscilloscopes normally have a transparent grid of 10 mm by 10 mm squares in front of the screen, called a graticule. Also on the front panel of a c. With no voltage applied to the Y plates the position of the spot trace on the screen is noted. Nag Materials Science by R. Khurmi and R. Sedha Thermodynamics for Chemists by S. Storey, E. Hughes and W. Balasubramaniam by R.
Avadhanulu, And P. Kshirsagar Energy Management by W. Murphy and G. Fox And A. Srivastava Electrical Machines by S. Kapoor Op-amp and linear integrated circuits by Dr. Mathur and R. Doolittle Mechanics of Structures by S.
Fundamental Electrical and Electronic Principles by C.R. Robertson
Junnarkar Modern Physics by G. Rajagopal Electronic Circuits by P. Gupta mechanics of fluid by B. Massey And A. Ward-Smith Integrated Circuits by P. Shaikh Electronic Instrumentation and Measurements by U. Shah Fiber Optics Communication by H. Kreith, R. Manglik And M. Bell Electronic Devices and Circuits by S.
Kakani and K. Bhandari Elements of Electric drives by J. Bell Engineering Physics by S. Srivastava and R. Yadav Applied Chemistry by Dr. Singal Electrical Power-i by M. Singh and D. Kulkarni Electronic Devices by K. Nandi Engineering Physics by S. Resnick Non-conventional Energy Sources by G. Wali Antenna and Wave Propagation by k.
Shinghal Solid State Electronics by J. Agrawal Engineering Physics by S. Jain and G. Sahasrabudhe Electronics Devices and Circuits by G. Raju Electronic Devices by S. Chaudhary and H. Maity Advanced Mechanics of Materials by A. Boresi and R. Verma Electrical Power System by C. Shukla Engineering Physics by P. Chandra And O. Crockford, Samuel B. Knight Introduction to Nuclear Engineering by J. Lamarsh and A. Prakash Thermodynamics by J. Jacob And G. Gupta Introduction to Electric Drives by J. Katre Electrical Network by R. Singh Optical Communication Systems by S.
Stevenson, Jr. Despande Elements of Mechanical Engineering by N. Bhatt and J. Esposito Elements of thermal technology by John H. Gupta, R. Manglik and R. Manglik Linear Integrated Circuits by J. Gupta Analog Integrated Circuits by J. Gupta Materials Science by Dr. Bakshi, A. Bakshi, K. Gupta Integrated Circuits by Dr.
Smith, H. Van Ness And M. Abbott Special Electrical Machines by S. Gupta Electronic Circuits by Dr. King Microelectronic Circuits by A. Sedra and K. Smith Quantum mechanics by M. Arnikar and N. Cathey Atomic and Nuclear Physics by N. Subrahmanyam, B. Lal And J. Seshan The Theory of Machines by T. Bevan Electronic Circuits by M. Tooley Turbomachines by A. Arasu High Voltage Engineering by C.
Mittal, R. Verma And S. Ghosh And A. Jeffords Internal Combustion Engines by H. Raghuvanshi Thermal Engineering by S. Askeland and P. Muthusubramanian and S. Salivahanan Applied Physics by K. Vijaya Kumar, T. Mishra Chemical Engineering Thermodynamics by T. Nandi Fluid Mechanics by A. Kuriacose And J. Johnston, D.
Mazurek, P. Cornwell And E. Eisenberg Principles of Communication Systems by H. Taub and D. Schilling Chemical Engineering Thermodynamics by S. Sundaram Heat and Thermodynamics by Brijlal and N. Subrahmanyam Optical Communication by R. Kasap Applied Thermodynamics and Engineering by T.
Eastop and A. Sharma Wireless Communications and Networking by V. Chandra Shekar, P. Sastry Electronics Engineering by P. Raja modern physics by Satish K. Beiser Engineering Physics volume - 2 by B. Pandey and S. Helfrick And W. J Pratap, G. Prakash Reddy, S. Asadullah, P Madhusudana Rao, B. Salam, H. Anis, A. El Morshedy and R. Radwan Microwave engineering by D. Pozar Engineering Physics Volume-2 by S. Gupta Linear Integrated Circuits by T. Chattopadhyay and P. Rakshit Machine Design by T. Wentzell, P. E Basic Electronics by A. Godse and U. Sen Principles of Electronics by V. Shultis and R.
Faw Basic mechanical engineering by Basant Agrawal , C. Doebelin And D. Manik Engineering Thermodynamics by Dr. Grewal principle of physics by V. Gobby Electronic Communication by D. Kumar And S. Jain Fundamentals Of Aerodynamics by J. Anderson Jr. Gupta Machine Design by U. Rajput basic electrical engineering by nagsarkar and sukhija A Textbook of Electronic Circuits by R. Sedha Power Plant Engineering by P. Salivahanan, N. Kumar And A. Vallavaraj Principle of Communication Engineering by A.
Singh and A. Chhabra Fiber Optics and Optoelectronics by R. Abulencia And L. Halder Discrete Mathematics by S. Lipschutz, M. Lipson And V. Chapra And R. Canale Electronic Devices and Circuits by J. Hambley Analog Electronics by U. Bakshi And A. Raina Electronics Circuits and Systems by Y. Kumar and M. Lathi Optical Fiber Communication by A. Selvarajan, S. Serway and J. Faughn Digital Communications by S. Haykin Linear Algebra by K.
Hoffman and R. Strang Numerical Methods by E. Balaguruswamy Principles of Physics by F. Arasu Stoichiometry by B. Bhatt And S. Shah and C. Kale Modern physics for engineers by S. Taneja Engineering Mechanics by A. Tayal Semiconductor circuit approximations by A. Malvino Electrical Machines - I by M.
Verma And V. Gupta Material Science by S. Kakani and A. Cengel Applied Chemistry by J. Dixon Engineering Mechanics by R. Vijaya Kumar, S. Glasstone Antenna and Wave Propogation by U. Bakshi and A. Balabhaskar, N. Srinivasa Rao, B. Hundy, A. Fundamentals of Fluid Mechanics by B. Elements of Electromagnetics by M. Fluidization Engineering by K. Industrial Instrumentation by K. Thermodynamics: A Core Course by R. Satellite Communication by Anil K.
Materials science and engineering an introduction by William D. Fundamentals of Electric Drives and Control by B. Thermodynamics by F P Durham by F. Engineering Physics by Dr. Vijaya Kumar. Problems In Fluid Flow by D. Introduction To Chemical Engineering by S. Chemical Engineering Thermodynamics by P. Stoichiometry And Process Calculations by K. Textbook of Engineering Chemistry by R.
Engineering Physics by D. Mechanics Of Fluids by A. Electronic devices and circuits by I. J Nagrath. Material Science In Engineering by Dr. Basic electrical and electronics engineering -1 by S K Bhattacharya and debasish de. Chemical Reaction Engineering by O. Engineering Physics by K. Chemical Engineering Thermodynamics by Y. Elements of electrical science by Mukopadhyay, Pant. Thyristors Theory And Applications by R.
Theory of Alternating Current Machinery by A. Elementary Fluid Mechanics by J. Theory Of Machines by R. Thermodynamics: From concepts to applications by A. Engineering Physics by H. Principles of Power System by V. Mehta, Rohit Mehta. How to think like a computer scientist by Allen B. Antennas and Wave Propagation by John. Engineering Physics - I by G. Fundamental of internal combustion engines by H N Gupta. Water and Wastewater Engineering by G. Principles Of Fluid Mechanics by M. Linear Integrated Circuits by D.
Introduction to Electrical Engineering by Er. Sonal Sapra. Concepts of Thermodynamics by Obert Edward F. Examples in Thermodynamics Problems by W. Mechanics of Materials by Ferdinand P. Electronic Measurements and Instrumentation by Er.
Principles of Electronics, 11/e
Thermodynamics Demystified by Merle C. Optical Fiber Communication by V. Elements of Physical Chemistry by Atkins Peter. Advance Semiconductor Devices by K. Fundamental Electrical and Electronic Principles by C. Discrete Mathematics and its Applications by Kenneth H. Mass - Transfer Operations by R.
Advanced Strength and Applied Elasticity by A. Power Electronics by Dr. B R Gupta, V Singhal. Aircraft Structures for Engineering Students by T. Applied Physics for Engineers by Neeraj Mehta. Elements of Thermodynamics and heat transfer by O. Physics for BSc Paper-3 by A. Optical Fiber Communication by R. Electric Machinery And Transformers by B.