#### DC Machines MCQs – New Circuit Model, Emf and Torque ( DC Machines ) MCQs

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**DC Machines MCQs – Circuit Model, Emf and Torque ( DC Machines ) MCQs**

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# EMF and Torque Production – 1

**1. In which mode machine is operating, given that conductor current is in the same direction of conductor emf?**

a) Motoring**b) Generating**

c) Can’t be determined using directions

d) In both modes for different cycles**Answer: b****Explanation: If the conductor current is in the same direction of conductor emf then machine outputs electrical power and absorbs mechanical power. So, when mechanical power is absorbed machine is said to be in a generating mode. When conductor emf and conductor current are in opposite directions then machine is said to be in a motoring mode.**

**2. Nature of the flux density wave in the air gap is__________ (for armature current equal to 0)****a) Flat topped with quarter wave symmetry**

b) Point topped with quarter wave symmetry

c) Flat topped with half wave symmetry

d) Point topped with half wave symmetry**Answer: a****Explanation: In a DC machine magnetic structure is such that the flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction.**

**3. In a DC machine, average energy stored in the magnetic field remains constant independent of the armature rotation.****a) True**

b) False**Answer: a****Explanation: In a DC machine, barring the irrecoverable losses of both electric and magnetic origin, there is balance between electrical and mechanical powers of the machine; the average energy stored in the magnetic field remains constant irrespective of armature rotation.**

**4. Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as__________****a) E1 = E2**

b) E1 > E2

c) E1 < E2

d) Can’t be determined**Answer: a****Explanation: In a DC machine flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction. Emf produced is independent of B-wave shape, thus we will get same value for both cases.**

**5. Average coil emf for 20 coil turns (E1) and 40 coil turns (E2), will have ratio E1/E2=____ (assuming all other parameters same for both machines).****a) 1/2**

b) 2/1

c) 1/4

d) 4/1**Answer: a****Explanation: Emf generated in a DC machine is directly proportional to number of coil turns, Flux per pole, number of poles and armature speed in rad/s. Thus, ratio E1/E2= 20/40 (assuming all other parameters same for both machines).**

**6. What is the average coil emf generated in a 4-pole DC machine having flux/pole equal to 0.1 wb rotating at 1500 rpm? (No. of coil sides = 100)**

a) 19 kV

b) 1.9 kV**c) 190 V**

d) 19 V**Answer: a****Explanation: Average coil emf generated= ∅ωNP/π.****E= 0.1*1500*100*4/3.14****E= 60000/3.14****E≅ 19 Kv.**

**7. Emf and torque produced in a DC machine are proportional to ________ and _________ respectively.**

a) Armature speed and armature emf

b) Armature emf and armature speed

c) Armature current and armature emf**d) Armature speed and armature current****Answer: d****Explanation: Average coil emf generated= ∅ωNP/π. Machine torque = ka*∅*Ia. Thus, average coil emf generated can also be represented as ka*∅*ω. So, average coil emf is directly proportional to ω (armature speed) and average torque is directly proportional to Ia (armature current).**

**8. What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?**

a) 2

b) 3**c) 2.5**

d) 4**Answer: c****Explanation: In an emf equation Nc= Cp * Np. Here, Cp= coils/ parallel path. Np is defined as number of turns per parallel paths which is also called as ratio of total armature conductors to the twice of number of parallel paths. Np= 10/(2*2)= 10/4= 2.5.**

**9.What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?****a) Bav*Ic*l*Zr**

b) Bav*Ic*l/Zr

c) Bav*Ic*Zr/l

d) Can’t be expressed**Answer: a****Explanation: Avg. conductor force f= Bav*l*Ic. Here, Bav= Average flux density over pole, l= acyive conductor length. Thus, torque T= Z*f = Bav*l*Ia*Z. This torque is constant because both the flux density wave and current distribution is fixed in space at all times.****T developed= Bav*Ic*l*Zr (Here, r= mean air gap radius).**

**10. What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?**

a) 0.785* 10^{-6}**b) 0.785* 10 ^{-3}**

c) 0.785* 10

^{-2}

d) 0.785* 10

^{-4}

**Answer: b**

**Explanation: Pole pitch is called as center to center distance between two adjacent poles. When measured in electrical degrees one pole itch is equal to 1800. Pole pitch can be calculated as ratio of 2πr/P.**

**Pole pitch= 2*3.14* 0.5* 10-3 / 4= 0.785* 10-3 m.**

**EMF and Torque Production – 2**

**1. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of flux/pole?**

a) 0.188 Wb

b) 18.88 Wb**c) 0.0188 Wb**

d) 1.888 Wb**Answer: c****Explanation: Flux is defined as flux density for a given surface area. Here, Surface area can be calculated and multiplied with B to give the value of flux. Flux= 2πr*l*B. now, for calculating flux per pole, divide it by P=4. So, Flux per pole after substituting all values is equal to 0.0188 Wb.**

**2. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of induced emf?****a) 188 V**

b) 276 V

c) 94 V

d) 188 mV**Answer: a****Explanation: Induced emf in a DC machine is equal to,**

**3. Coil torque for 20 kA armature current (T1) and 40 mA armature current (T2), will have ratio T1/T2=____ (assuming all other parameters same for both machines).****a) 1/2**

b) 2/1

c) 1/4

d) 4/1**Answer: a****Explanation: Torque produced in a DC machine is directly proportional to number of coil turns, Flux per pole, number of poles and armature current. Thus, ratio T1/T2= 20/40 (assuming all other parameters same for both machines).**

**4. If the no load speed of DC motor is 1300 rpm and full load speed is 1100 rpm, then its voltage regulation is ____________**

a. 12.56%**b. 18.18 %**

c. 17.39%

d. 18.39%**Answer: b****Explanation: For A DC machine when all other parameters are fixed average coil emf generated is proportional directly to the speed of the dc motor. Voltage regulation is defined as ratio of difference of no load voltage and full load voltage to the full load voltage. VR= (1300-1100/1100)*100%.**

**5. If the average coil emf of a DC motor is doubled and flux is halved (keeping other parameters constant) then its shaft speed will become ___________**

a. Twice of the original speed

b. Square of the original speed**c. Four times of the original speed**

d. Half of the original speed**Answer: c****Explanation: Induced emf in a DC machine is equal to,****From the emf equation we get speed of the shaft i.e. n α E/Z, when all other parameters are kept constant. So, when E is doubled n becomes twice the original, halving flux on reduced emf will quadrupled the speed of a DC machine.**

**6. A 4-pole wave wound DC motor drawing an armature current of 20 A has provided with 360 armature conductors. If the flux per pole is 0.015 Wb then the torque developed by the armature of motor is _______**

a. 10.23 N-m**b. 34.37 N-m**

c. 17.17 N-m

d. 19.08 N-m**Answer: b****Explanation: DC Machine torque equation: T = ka*∅*Ia. Here, ka= ZP/(2πA), Z= total armature conductors, P= No. of poles, A= No. of parallel paths. For a wave winding A=2. So, substituting all the values in the torque equation we get torque equal to 34.37 N-m.**

**7. In a DC machine, what is the torque induced beyond the pole shoes?****a) 0**

b) 2/π *∅*i

c) π *∅*i/2

d) Can’t be calculated**Answer: a****Explanation: In a construction of a DC machine poles are located in magnetically neutral region. The magnetic field at the pole terminals in a DC machine will be equal to 0. Thus, cross product with the current flowing through the armature yields zero.**

**8. For a constant emf, if field current is reduced then the speed of the DC motor will_____**

a) Remains same**b) Increases**

c) Decreases

d) Can’t say**Answer: b****Explanation: When the field current is reduced, the field produced by the field winding also reduces. Thus, the term Φ from the emf equation also decreases. For all other parameters kept constant speed of the DC machine is inversely proportional to field. Hence, speed of DC motor will increase.**

**9. For an ideal DC machine, which phenomenon will reduce the terminal voltage?**

a) Armature reaction**b) Commutation**

c) Armature ohmic losses

d) All will contribute in reducing the terminal voltage**Answer: b****Explanation: In an ideal case, Commutation does not reduce the terminal voltage of a dc machine. In a non-ideal case, commutation takes place improperly at desired timings, thus losses contribute to the terminal voltage reduction. Armature reaction, ohmic losses due to winding resistance contribute to the losses in the terminal voltage.**

**Electromagnetic Power and Circuit Models**

**1. Product of torque and mechanical angular velocity ω is_____**

a) Ea/ω**b) Ea*Ia**

c) ω/Ea

d) Can’t tell**Answer: b****Explanation: According to statement of energy conversion, electrical and mechanical power of the machine must balance in a machine. Ea*Ia is referred to as electromagnetic power. Thus, torque = (Electromagnetic power)/ ω.**

**2. Condition for linear magnetization is______****a) φ α I _{f}**

b) φ α Ia

c) φ α 1/I

_{f}

d) φ α 1/Ia

**Answer: a**

**Explanation: For a linear magnestization to take place, flux produced by the current flowing through particaluar flux producing coil must vay in direct proportion. Here, φ is produced due to field winding which carried current I**

_{f}.**3. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. Here, motor is drawing 25 A current at 1500 rpm.****What is the available torque (N-m) at shaft?****a) 29.9**

b) 59.8

c) 14.95

d) 44.85**Answer: a****Explanation: Gross mechanical power developed is equal to Ea*Ia. Ea*Ia is referred to as electromagnetic power. Thus, torque = (Electromagnetic power)/ ω. Substituting all the values we get torque = 29.9 Nm.**

**4.When Ea>Vt machine is said to be operating in which of the following mode?**

a) Depends on the Shaft speed**b) Generating**

c) Motoring

d) Inducing**Answer: b****Explanation: The machine operates in generating mode (puts out electrical power) when Ia is in the direction of induced emf Ea. For the given armature circuit Vt = armature terminal voltage= Ea – Ia * Ra; which implies Vt < Ea.**

**5. When Ea<Vt machine is said to be operating in which of the following mode?**

a) Depends on the Shaft speed

b) Generating**c) Motoring**

d) Inducing**Answer: c****Explanation: The machine operates in motoring mode (puts in electrical power) when Ia is in the direction opposite of induced emf Ea. For the given armature circuit Vt = armature terminal voltage= Ea + Ia * Ra; which implies Vt> Ea.**

**6. Simple equation of DC machine operating in generating mode, with non-zero Ra value is________**

a) Vt = Ea – Ia/Ra

b) Vt = Ea + Ia*Ra

c) Vt = – Ea + Ia*Ra**d) Vt = Ea – Ia*Ra****Answer: d****Explanation: The machine operates in generating mode (puts out electrical power) when Ia is in the direction of induced emf Ea, also Ea>Vt. For the given armature circuit, with non-zero value of Ra, Vt= armature terminal voltage= Ea- Ia*Ra.**

**7. Simple equation of DC machine operating in motoring mode, with non-zero Ra value is________**

a) Vt = Ea – Ia/Ra**b) Vt = Ea + Ia*Ra**

c) Vt = – Ea + Ia*Ra

d) Vt = Ea – Ia*Ra**Answer: b****Explanation: The machine operates in motoring mode (puts in electrical power) when Ia is in the direction opposite of induced emf Ea, also Ea<Vt. For the given armature circuit, with non-zero value of Ra, Vt= armature terminal voltage= Ea+ Ia*Ra.**

**8. When torque of the electromagnetic origin is in the opposite direction of rotation of armature, machine is said to be operating in which of the following mode?**

a) Depends on other parameters**b) Generating**

c) Motoring

d) Inducing**Answer: b****Explanation: In a generating mode, torque of the electromagnetic origin is in the opposite direction of rotation of armature, implying mechanical power is absorbed and a prime mover is needed to run the machine. Ea and Ia are in the same directions.**

**9. When torque of the electromagnetic origin is in the direction of rotation of armature, machine is said to be operating in which of the following mode?**

a) Depends on other parameters

b) Generating**c) Motoring**

d) Inducing**Answer: c****Explanation: In a motoring mode, torque of the electromagnetic origin is in the direction of rotation of armature, implying electrical power is absorbed and a prime mover is not needed to run the machine. Ea and Ia are in the opposite directions.**

**10. What is the armature current for a DC motor if terminal voltage is 255V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?**

a) 10A**b) 100A**

c) 1KA

d) 1A**Answer: b****Explanation: Open circuit voltage means the voltage at armature current equal to 0, i.e. Vt=Ea= 250V. Actual Vt at loaded condition is 255V. For motoring mode, Vt – Ea = Ia*Ra. By substituting all the given values, we get Ia= 100A.**

**11. For a DC generator feeding 100kW power into 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____****a) 436.8 A**

b) 434.8 a

c) 432.8 A

d) Data insufficient**Answer: a****Explanation: Field current is equal to 230/115 = 2A, by Ohm’s law. When running as a generator line current IL= 100k/230= 434.8A. Since power is supplied to 230V mains, Ia=Il+If = 434.8+2 =436.8A.****For motoring mode Ia= IL- If.**

**12. What is the armature current for a DC generator if terminal voltage is observed to be 245V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?**

a) 10A**b) 100A**

c) 1KA

d) 1A**Answer: b****Explanation: Open circuit voltage means the voltage at armature current equal to 0, i.e. Vt=Ea= 250V. Actual Vt at loaded condition is 245V. For motoring mode, Ea – Vt= Ia*Ra. By substituting all the given values, we get Ia= 100A.**

**13. For a DC motor taking 10kW power from 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____**

a) 45.68 A

b) 43.48 a**c) 41.48 A**

d) Data insufficient**Answer: c****Explanation: Field current is equal to 230/115 = 2A, by Ohm’s law. When running as a motor line current IL= 10k/230= 43.48A. Since power is taken from 230V mains, Ia=IL- If = 43.48-2 =41.48A. For generating mode Ia= IL+ If.**