Unit 4 DC Generator PDF

# Principles of Electrical Engineering ## 482 Principles of Operation * The operation of all rotating electrical machines are based on the following two fundamental electromagnetic laws: * **When the magnetic flux linking a conductor or coil changes, an e.m.f. is induced in it whose magnitude...

# Principles of Electrical Engineering ## 482 Principles of Operation * The operation of all rotating electrical machines are based on the following two fundamental electromagnetic laws: * **When the magnetic flux linking a conductor or coil changes, an e.m.f. is induced in it whose magnitude is given by:** * $e = \frac{Ndo}{dt}$ ...Faraday's Law * This is the basic equation that governs the induced voltage in the armature winding whether the electrical machine is d.c. or a.c. The ultimate forms of the induced voltage equations appear different for the two types of machines only because the details of mechanical construction differ. * **When a current-carrying conductor of length I is placed at right angles to a uniform magnetic field (of flux density B), the conductor experiences a mechanical force (F) whose magnitude is given by:** * $F = Bil$ ...Maxwell's Law * This basic equation of force is used to find the developed torque (T) whether the electrical machine is d.c. or a.c. The ultimate forms of the torque equations appear different for the two types of machines only because the details of mechanical construction differ. ## 12.6 D.C. Machines * Although a far greater percentage of the electrical machines in service are a.c. machines, the d.c. machines are of considerable industrial importance. * The principal advantage of the d.c. machine, particularly the d.c. motor, is that it provides a fine control of speed. * Such an advantage is not claimed by any a.c. motor. However, d.c. generators are not as common as they used to be, because direct current, when required, is mainly obtained from an a.c. supply by the use of rectifiers. * Nevertheless, an understanding of d.c. generator is important because it represents a logical introduction to the behaviour of d.c. motors. * Indeed many d.c. motors in industry actually operate as d.c. generators for a brief period. * We first discuss d.c. generators followed by d.c. motors. ## 12.7 Generator Principle * An electric generator is a machine that converts mechanical energy into electrical energy. * An electric generator is based on the principle that whenever magnetic flux is cut by a conductor, an e.m.f. is induced which will cause a current to flow if the conductor circuit is closed. * The direction of induced e.m.f. (and hence current) is given by *Fleming's Right hand rule. Therefore, the essential components of a generator are : * a magnetic field * conductor or a group of conductors * motion of conductor w.r.t. magnetic field ## 12.8 Simple Loop Generator * Consider a single turn loop ABCD rotating clockwise in a uniform magnetic field with a constant speed as shown in Fig. 12.4. * As the loop rotates, the magnetic flux linking the coil sides AB and CD changes continuously. * Hence the e.m.f. induced in these coil sides also changes but the e.m.f. induced in one coil side **adds to that induced in the other.** * When the loop is in position no. 1 [See Fig. 12.4], the generated e.m.f. is zero because the coil sides (AB and CD) are cutting no magnetic flux but are moving parallel to it. * **Fleming's Right hand rule.** Stretch the thumb, fore-finger and middle finger of your right hand so that they are at right angles to each other. If the fore-finger points in the direction of field, thumb in the direction of motion of the conductor, then middle finger will point in the direction of induced e.m.f. * It is because the coil sides always remain under the influence of opposite poles i.e. if one coil side is under the influence of the N-pole, then the other coil side will be under the influence of S-pole and vice-versa. * When the loop is in position no. 2, the coil sides are moving at an angle to the magnetic flux and, therefore, a low e.m.f. is generated as indicated by point 2 in Fig. 12.5. * When the loop is in position no. 3, the coil sides (AB and CD) are at right angle to the magnetic flux and are, therefore, cutting the flux at a maximum rate. Hence at this instant, the generated e.m.f. is maximum as indicated by point 3 in Fig. 12.5. * At position 4, the generated e.m.f. is less because the coil sides are cutting the magnetic flux at an angle. * At position 5, no magnetic lines are cut and hence, induced e.m.f. is zero as indicated by point 5 in Fig. 12.5. * At position 6, the coil sides move under a pole of opposite polarity and hence the direction of generated e.m.f. is reversed. The maximum e.m.f. in this direction (i.e. reverse direction, See Fig. 12.5) will be when the loop is at position 7 and zero when at position 1. This cycle repeats with each revolution of the coil. * Note that e.m.f. generated in the loop is alternating one. It is because any coil side, say AB, has e.m.f. in one direction when under the influence of N-pole and in the other direction when under the influence of S-pole. If a load is connected across the ends of the loop, then alternating current will flow through the load. The alternating voltage generated in the loop can be converted into direct voltage by a device called commutator. We then have the d.c. generator. In fact, a commutator is a mechanical rectifier. ## 12.9 Action of Commutator * If, somehow, connection of the coil side to the external load is reversed at the same instant the current in the coil side reverses, the current through the load will be direct current. This is what a commutator does. * Fig. 12.6 shows a commutator having two segments C₁ and C. It consists of a cylindrical metal ring cut into two halves or segments C₁ and C₂ Mica respectively separated by a thin sheet of mica. The commutator is mounted on, but insulated from the rotor shaft. The ends of coil sides AB and CD are connected to the segments C₁ and C2 respectively as shown in Fig. 12.7. Two stationary carbon brushes rest on the commutator and lead current to the external load. * With this arrangement, the commutator at all times connects the coil side under S-pole to the +ve brush and that under N-pole to the -ve brush. * In Fig. 12.7, the coil sides AB and CD are under N-pole and S-pole respectively. Note that segment C₁ connects the coil side AB to point P of the load resistance R and the segment C₂ connects the coil side CD to point Q of the load. Also note the direction of current through load. It is from Q to P. * After half a revolution of the loop (i.e. 180° rotation), the coil side AB is under S-pole and the coil side CD under N-pole as shown in Fig. 12.8. The currents in the coil sides now flow in the reverse direction but the segments C₁ and C₂ have also moved through 180° i.e. segment C₁ is now in contact with + ve brush and segment C₂ in contact with ve brush. Note that commutator has reversed the coil connections to the load i.e. coil side AB is now connected to point Q of the load and coil side CD to the point P of the load. Also note the direction of current through the load. It is *again from Q to P. * Thus the alternating voltage generated in the loop will appear as direct voltage across the brushes. The reader may note that e.m.f. generated in the armature winding of a d.c. generator is alternating one. It is by the use of commutator that we convert the generated alternating e.m.f. into direct voltage. The purpose of brushes is simply to lead current from the rotating loop or winding to the external stationary load. * The variation of voltage across the brushes with the angular displacement of the loop will be as shown in Fig. 12.9. This is not a steady direct voltage but has a pulsating character. It is because the voltage appearing across the brushes varies from zero to maximum value and back to zero twice for each revolution of the loop. A pulsating direct voltage such as is produced by a single loop is not suitable for many commercial uses. What we require is the steady direct voltage. This can be achieved by using a large number of coils connected in series. The resulting arrangement is known as armature winding. * The situation is that currents in the coil sides are reversed and at the same time the connections of the coil sides to the external load are reversed. This means that current will flow in the same direction through the load. ## 12.10 Construction of D.C. Generator (or Motor) * The d.c. generators and d.c. motors have the same general construction. In fact, when the machine is being assembled, the workmen usually do not know whether it is a d.c. generator or motor. Any d.c. generator can be run as a d.c. motor and vice-versa. All d.c. generator principal components viz (i) field system (ii) armature core (iii) armature winding (iv) commutator (v) brushes [See Fig. 12.10]. * **Field system.** The function of the field system is to produce uniform magnetic field within which the armature rotates. It consists of a number of salient poles (of course, even number) bolted to the inside of circular frame (generally called yoke). The yoke is usually made of solid cast steel whereas the pole pieces are composed of stacked laminations. Field coils are mounted on the poles and carry the d.c. exciting current. The field coils are connected in such a way that adjacent poles have opposite polarity. The m.m.f. developed by the field coils produces a magnetic flux that passes through the pole pieces, the air gap, the armature and the frame (See Fig. 12.11). Practical d.c. machines have air gaps ranging from 0.5 mm to 1.5 mm. Since armature and field systems are composed of materials that have high permeability, most of the m.m.f. of field coils is required to set up magnetic flux in the air gap. By reducing the length of air gap, we can reduce the size of field coils (i.e. number of turns). * **Armature core.** The armature core is keyed to the machine shaft and rotates between the field poles. It consists of slotted soft-iron laminations (about 0.4 to 0.6 mm thick) that are stacked to form a cylindrical core as shown in Fig. 12.12. The laminations (See Fig. 12.13) are individually coated with a thin insulating film so that they do not come in electrical contact with each other. The purpose of laminating the core is to reduce the eddy current loss. The laminations are slotted to accommodate and provide mechanical security to the armature winding and to give shorter air gap for the flux to cross between the pole face and the armature "teeth". * **Armature winding.** The slots of the armature core hold insulated conductors that are connected in a suitable manner. This is known as armature winding. This is the winding in which “working" e.m.f. is induced. The armature conductors are connected in series-parallel; the conductors being connected in series so as to increase the voltage and in parallel paths so as to increase the current. The armature winding of a d.c. machine is a closed-circuit winding; the conductors being connected in a symmetrical manner forming a closed loop or series of closed loops. * **Commutator.** A commutator is a mechanical rectifier which converts the alternating voltage generated in the armature winding into direct voltage across the brushes. The commutator is made of copper segments insulated from each other by mica sheets and mounted on the shaft of the machine (See Fig. 12.14). The armature conductors are soldered to the commutator segments in a suitable manner to give rise to the armature winding. Depending upon the manner in which the armature conductors are connected to the commutator segments, there are two types of armature winding in a d.c. machine viz (a) lap winding (b) wave winding. * Great care is taken in building the commutator because any eccentricity will cause the brushes to bounce, producing unacceptable sparking. The sparks may burn the brushes and overheat and carbonise the commutator. * **Brushes.** The purpose of brushes is to ensure *electrical connections between the rotating commutator and stationary external load circuit. The brushes are made of ** carbon and rest on the commutator. The brush pressure is adjusted by means of adjustable springs (See Fig. 12.15). If the brush pressure is very large, the friction produces heating of the commutator and the brushes. On the other hand, if it is too weak, the imperfect contact with the commutator may produce sparking. ## 12-11. Types of Armature Windings * The armature windings of d.c. machines are always of drum type. In this arrangement, the armature conductors, usually in the form of coils, are placed in slots around the complete surface of drum-shaped or cylindrical armature core. The coils are connected in series through the commutator segments in such a way that their e.m.f.s add to each other. There are two types of d.c. armature windings viz (i) Wave winding (ii) Lap winding. * **Wave winding.** In this arrangement, the armature coils are connected in series through commutator segments in such a way that the armature winding is divided into two parallel paths irrespective of the number of poles of the machine. If there are Z armature conductors, then Z/2 conductors will be in series in each parallel path as shown in Fig. 12.16. Each parallel path will carry a current 1/2 where I is the total armature current. * **Lap winding.** In this arrangement, the armature coils are connected in series through commutator segments in such a way that the armature winding is divided into as many parallel paths as the number of poles of the machine. If there are Z conductors and p poles, then there will be P parallel paths, each containing ZIP conductors in series as shown in Fig. 12.17. Each parallel path will carry a current of I/P where I is the total armature current. Here it is assumed that P = 4 so that there are 4 parallel paths. * To sum up, in a simple wave winding : * there are two parallel paths irrespective of number of poles of the machine. * each parallel path has 2/2 conductors in series; Z being total number of armature conductors. * e.m.f. generated = e.m.f./parallel path * total armature current, I = 2 x current/parallel path * To sum up, in a simple lap winding : * there are as many parallel paths as the number of poles (P) of the machine. * each parallel path has Z/P conductors in series where Z and P are the total number of armature conductors and poles respectively. * e.m.f. generated = e.m.f./parallel path. * total armature current, I ₁ = P × current/parallel path. * **Applications of lap and wave windings.** In multipolar machines, for a given number of poles (P) and armature conductors (Z), a wave winding has a higher terminal voltage than a lap winding because it has more conductors in series. On the other hand, the lap winding carries more current than a wave winding because it has more parallel paths. In small machines, the current-carrying capacity of the armature conductors is not critical and in order to achieve suitable voltages, wave windings are used. On the other hand, in large machines suitable voltages are easily obtained because of the availability of large number of armature conductors and the current carrying capacity is more critical. Hence, in large machines, lap windings are used. **Note.** In general, a high-current armature is lap-wound to provide a large number of parallel paths and a low-current armature is wave-wound to provide a small number of parallel paths. ## 12.12. E.M.F. Equation of a D.C. Generator * We shall now derive an expression for the e.m.f. generated in a d.c. generator. * Let * $Φ$ = magnetic flux/pole in Wb * Z = total number of armature conductors * P = number of poles * A = number of parallel paths = 2 ....for wave winding * $A = \frac{P}{2}$ ....for lap winding * N = speed of armature in r.p.m. * E = e.m.f. of the generator * Magnetic flux cut by one conductor in one revolution of the armature is $dΦ = PΦ$ webers * Time taken to complete one revolution is $dt = \frac{60}{N}$ second * e.m.f. generated/conductor= $\frac{dΦ}{dt} = \frac{PΦ}{60/N} = \frac{PΦN}{60}$ volts * e.m.f. of generator, $E = (e.m.f./conductor) x no.$ of conductors in series per parallel path * $E = \frac{PΦN}{60} × \frac{Z}{A}$ * $E = \frac{PΦZN}{60A}$ ...(i) * where * $A = 2$ ....for wave winding * $A = \frac{P}{2}$ ....for lap winding * **Discussion.** The following points may be noted : * Eq. (i) is also valid for a.d.c. motor. In case of a d.c. generator, induced e.m.f. in the armature is called generated e.m.f. or generated voltage (E). When the d.c. machine operates as a motor, the induced e.m.f. in the armature is called counter e.m.f. or back e.m.f. $E_g$ (= POZN/60A). * For a given d.c. machine (generator or motor), Z, P and A are constant so that $E_g$ (or $E_b$) ∝ $ΦN$. Therefore, for a given d.c. machine, the induced voltage in the armature is directly proportional to flux per pole (Φ) and speed of rotation (N). * **Note.** In deriving eq. (i) above, we have assumed that the poles cover the entire armature periphery. This is virtually impossible. In practical machines, the poles cover 60% to 80% of the armature periphery. This fact must be taken into consideration while computing flux per pole (Φ). * **Example 12.1.** Derive e.m.f. equation of a d.c. generator. What will be the change in e.m.f. induced if the flux is reduced by 20% and the speed is increased by 20%? * **Solution.** Induced e.m.f. ∝ $ΦN$ * **or** $\frac{E_2}{E_1} = \frac{Φ_2N_2}{Φ_1N_1} = \frac{Φ_2}{Φ_1} × \frac{N_2}{N_1} = 0.8 × 1.2 = 0.96$ * $E_2 = 0.96 E_1$ * **or** $\frac{E_1 - E_2}{E_1} × 100 = \frac{E_1 - 0.96 E_1}{E_1} × 100 = 4%$ decrease * **Change in e.m.f. = 4% decrease** * **Example 12.2.** A 6-pole lap-wound d.c. generator has 600 conductors on its armature. The flux per pole is 0.02 Wb. Calculate (i) the speed at which the generator must be run to generate 300V. (ii) What would be the speed if the generator were wave-wound? * **Solution.** (i) Lap wound. For lap-wound generator, number of parallel paths, A = P. * **Now,** * $E_g = \frac{PΦZN}{60A}$ * $N = \frac{E_g × 60A}{PΦZ} = 800 r.p.m.$ * **(ii) Wave wound.** For wave-wound generator, number of parallel paths, A = 2. * $N = \frac{E_g × 60A}{PΦZ} = 500 r.p.m.$ * **Example 12.3.** A d.c. generator has an armature voltage of 100 V when the useful flux per pole is 20 mWb and the speed is 800 r.p.m. Calculate the generated e.m.f. (i) with the same flux and a speed of 1000 r.p.m. (ii) with a flux per pole of 24 mWb and a speed of 900 r.p.m. * **Solution.** $E_1$ = 100 V; $Φ_1$ = 20 mWb = 20 × 10⁻³ Wb; N₁ = 800 r.p.m * (i) $E_2 = \frac{Φ_2N_2}{Φ_1N_1} × E_1 = 100 × \frac{1000}{800} = 125 V$ * (ii) $E_3 = \frac{Φ_3N_3}{Φ_1N_1} × E_1 = 100 × \frac{24 × 10⁻³ × 900}{20 × 10⁻³ × 800} = 135 V$ * **Example 12.4.** An 8-pole, lap-wound armature rotated at 350 r.p.m. is required to generate 260 V. The useful magnetic flux per pole is 0.05 Wb. If the armature has 120 slots, calculate the number of conductors per slot. * **Solution.** * $E_g = \frac{PΦZN}{60A}$ * $Z = \frac{E_g × 60A}{PΦN} = 890$ * No. of conductors/slot = $\frac{890}{120}$ = 7.14 * This value must be an even number. * Hence, conductors/slot = 8 * **Example 12.5.** The armature of a 6-pole, 600 r.p.m. lap-wound generator has 90 slots. If each coil has 4 turns, calculate the flux per pole required to generate an e.m.f. of 288 volts. * **Solution.** Each turn has two active conductors and 90 coils are required to fill 90 slots. * $Z = 90 × 4 × 2 = 720$ * $E_g = \frac{PΦZN}{60A}$ * $Φ = \frac{E_g × 60A}{PZN} = 0.04 Wb$ * **Example 12.6.** The armature of a d.c. generator has 81 slots and the commutator has 243 segments. It is wound to give lap winding having 1 turn per coil. If the flux per pole is 30 mWb, calculate the generated e.m.f. at a speed of 1200 r.p.m. Number of poles = 6. * **Solution.** The number of coils is equal to the number of commutator segments. Each turn has 2 active conductors. * $Z = 243 × 2 = 486$ * $E_g = \frac{PΦZN}{60A} = 291.6 volts$ * **Example 12.7.** A generator has a rated armature current of 180A. What is the current per path of the armature if the armature is (i) simplex wave wound (ii) simplex lap wound? The machine has 12 poles. * **Solution.** * Armature current, $I_a$ = 180A; P = 12 poles. * (i) For simplex wave wound armature, number of parallel paths, A = 2. * Current per parallel path = $I_a/A = 90 A$ * (ii) For simplex lap wound armature, number of parallel paths, A = P. * Current per parallel path = $I_a/A = 15 A$ ## TUTORIAL PROBLEMS 1. An 8-pole d.c. generator has 500 armature conductors and a useful flux of 0.05 Wb. What will be the e.m.f. generated, if it is lap-connected and runs at 1200 r.p.m.? What must be the speed at which it is to be driven to produce the same e.m.f., if it is wave-wound ? [500V; 300 r.p.m] 2. A lap-connected 8-pole generator has 500 armature conductors and useful flux per pole of 0.07Wb. Determine the induced e.m.f. when it runs at 1000 r.p.m. What must be the speed at which it is to be driven to produce the same e.m.f. if it is wave wound? [583.33V; 250 r.p.m] 3. A lap wound d.c. generator having 8-poles develops an e.m.f. of 500V at 400 r.p.m. The armature has 144 slots and each slot contains 6 conductors. Calculate the flux per pole. [0.0868 Wb] ## 12-13. Armature Resistance (R_a) * The resistance offered by the armature circuit is known as armature resistance (R_a) and includes : * resistance of armature winding * resistance of brushes * The armature resistance depends upon the construction of machine. Except for small machines, its value is generally less than 12. ## 12-14. Types of D.C. Generators * The magnetic field in a d.c. generator is normally produced by electromagnets rather than permanent magnets. Generators are generally classified according to their methods of field excitation. On this basis, d.c. generators are divided into the following two classes : * Separately excited d.c. generators * Self-excited d.c. generators * The behaviour of a d.c. generator on load depends upon the method of field excitation adopted. ## 12.15. Separately Excited D.C. Generators * A d.c. generator whose field magnet winding is supplied from an independent external d.c. source (e.g. a battery etc.) is called a separately excited generator. Fig. 12.18 shows the connections of a separately excited generator. The voltage output depends upon the speed of rotation of armature and the field current ($E_g = PΦZN/60A$). * The greater the speed and field current, greater is the generated e.m.f. It may be noted that separately excited d.c. generators are rarely used in practice. The d.c. generators are normally of self-excited type. ## 12-16. Self-Excited D.C. Generators * A d.c. generator whose field magnet winding is supplied current from the output of the generator itself is called a *self-excited generator*. There are three types of self-excited generators depending upon the manner in which the field winding is connected to the armature, namely; * Series generator (ii) Shunt generator (iii) Compound generator * **Series generator.** In a series-wound generator, the field winding is connected in series with armature winding so that whole armature current flows through the field winding as well as the load. Fig. 12.19 shows the connections of a series-wound generator. Since the field winding carries the whole of load current, it has a few turns of thick wire having low resistance. Series generators are rarely used except for special purposes e.g. as boosters. * **Shunt generator.** In a shunt generator, the field winding is connected in parallel with the armature winding so that terminal voltage of the generator is applied across it. The shunt field winding has many turns of fine wire having high resistance. Therefore, only a part of armature current flows through shunt field winding and the rest flows through the load. Fig. 12.20 shows the connections of a shunt-wound generator. * **Compound generator.** In a compound-wound generator, there are **two sets of field windings on each pole-one is in series and the other in parallel with the armature. A compound wound generator may be: * *How is self-excitation achieved?* When the armature is rotated, a small voltage is induced in the armature winding due to residual magnetic flux in the poles. This voltage produces a small field current in the field winding and causes the flux per pole to increase. The increased flux increases the induced voltage which further increases the field current. These events take place rapidly and the generator builds up to the rated generated voltage. * *Normally, the majority of m.m.f. is provided by the shunt field. The two windings may be connected to aid each other (cumulative compounding) or they may oppose each other (differential compounding)* * (a) Short Shunt in which only shunt field winding is in parallel with the armature winding [See Fig. 12.21 (i)]. * (b) Long Shunt in which shunt field winding is in parallel with both series field and armature winding [See Fig. 12.21 (ii)]. * **Short shunt.** * Series field current, $I_{se} = I_L$ * Terminal voltage, $V = E_g - I_aR_a - I_{se}R_{se}$ * Power developed in armature = $E_gI_a$ * Power delivered to load = $VI_L$ * **Long shunt.** * Series field current, I * Terminal voltage, $V = E_g - I_a (R_a + R_{se})$ * Power developed in armature = $E_gI_a$ * Power delivered to load = $VI_L$ ## 12-17. Brush Contact Drop * It is the voltage drop over the brush contact resistance when current flows. Obviously, its value will depend upon the amount of current flowing and the value of contact resistance. This drop is generally small and may be neglected if not given. * **Example 12.8.** A 4-pole, lap-connected d.c. machine has an armature resistance of 0.15 Ω. Find the armature resistance of the machine if rewound for wave connection. * **Solution.** Let r ohm be the resistance of each parallel path. Then r/4 = 0.15. Therefore, r = 0.6Ω. When the machine is rewound for wave connection, then there will be two parallel paths each of resistance = 2r = 2 × 0.6 = 1.2 Ω. Therefore, armature resistance for wave wound machine = 1.2||1.2 = 1.2/2 = 0.6 Ω. * **Example 12.9.** The armature of a d.c. generator consists of 40 coils and each coil has 20 turns. When the armature is rotated at 200 rad/s in a 4-pole field structure having a flux of 5 mWb/pole and there are four paths in the armature, calculate (i) number of armature conductors (ii) the voltage between brushes generated by the armature. * **Solution.** (i) Number of armature conductors is = 1600 * *Each turn has 2 conductors.* * (ii) Voltage between brushes, $E_g = \frac{PΦΖω}{2πA} = 254.6 V$ * **Example 12.10.** A 4-pole d.c. shunt generator with a wave-wound armature has to supply a load of 500 lamps each of 100 W at 250 V. Allowing 10 V for voltage drop in the connecting leads between the generator and the load and drop of 1V per brush, calculate the speed at which the generator should be driven. The flux per pole is 30 mWb and the armature and shunt field resistances are respectively 0.05 Ω. and 65 2. The number of armature conductors is 390. * **Solution.** * Load current, $I_L = 200 A$ * Voltage across shunt, $I_{sh} = 4 A$ * $I_a = I_L + I_{sh} = 204 A$ * Brush drop = 2V * $E_g = 260 + 2 + 204 × 0.05 = 272.2 V$ * $E_g = \frac{PΦZN}{60A}$ * $N = \frac{E_g × 60A}{PΦZ} = 698 r.p.m.$ * **Example 12.11.** A 30 kW, 300 V d.c. shunt generator has armature and field resistances of 0.05 Ω and 100 Ω respectively. Calculate the total power developed by the armature when it delivers full load output. * **Solution.** Fig. 12.22 shows the shunt generator circuit. * $I_L = 100 A $ * $I_{sh} = 3A$ * $I_a = I_L + I_{sh} = 103 A$ * $E_g = V + I_aR_a = 305.15 V$ * Power developed by armature = $E_gI_a = 31.43 kW$ ## 12.12. A 4–pole lap-wound d.c. shunt generator has a useful flux per pole of 0.07 Wb. The armature winding consists of 220 turns, each of 0.004 resistance. Calculate the terminal voltage when running at 900 r.p.m. if the armature current is 50 A. * **Solution.** * $E_g = \frac{PΦZN}{60A} = 462 V $ * No. of turns per parallel path = 55 * Resistance per parallel path = 0.055 Ω * $E_g = V+IR_a$ * **V = 459.25 volts** ## 12.13. A d.c. shunt generator runs at 400 r.p.m.and delivers 500 kW to busbars having a constant voltage of 400V. Assuming the field excitation to be constant at 5A, calculate the speed at which the generator must run if the load on it is to be reduced to 300 kW. The armature resistance is 0.015 Ω. * **Solution.** * In the present problem, Φ is given to be constant. Therefore, $E_g ∝ N$. * **Initial conditions** * Load current, I₁ = 1250 A * Armature current, I