The Direct-Current Motor Problems and Solutions
Problem#1
A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 Ω. When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?Answer:
Known:
internal resistance, r = 3.2 Ω
emf in the rotor, ε = 105 V
potential difference, Vab = 120 V
(a) the current drawn by the motor from the line is
Vab = ε + Ir
I = (Vab – ε)/r = (120 V – 105 V)/3.2 Ω = 4.7 A
(b) the power delivered to the motor is
Psupplied = IVab = 4.7 A x 120 V = 564 W
(c) the mechanical power developed by the motor is
Pmech = IVab – I2r = 564 W – (4.7 A)2 x 3.2 Ω = 493 W
Problem#2
In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig.1), the resistance Rf of the field coils 106 Ω is and the resistance Rr of the rotor is 5.9 Ω. When a potential difference of 120 V is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.82 A. (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?
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| Fig.1 |
Answer:
(a) the current in the field coils is
Vf = IfR
If = Vf/R = 120 V/106 Ω = 1.13 A
(b) the current in the rotor is
I – If – Ir = 0 (applying the juction rule to point a in the circuit diagram)
Ir = I – If = 4.82 A – 1.13 A = 3.69 A
(c) the induced emf developed by the motor is
ε = V – IrRr = 120 V – 3.69 A x 5.9 Ω = 98.2 V
(d) The mechanical power output is the electrical power input minus the rate of dissipation of electrical energy in the resistance of the motor:
Electrical power input to the rotor
Pin = IV = (4.82 A)(120 V) = 578 W
Electrical power loss in the to resistances
Ploss = If2Rf + Ir2R
Ploss = (1.13 A)2(106 Ω) + (3.69 A)2(5.9 Ω) = 216 W
Mechanical power output
Pout = Pin – Ploss = 578 W – 216 W = 362 W
Or the mechanical power output is the power assosiated with the incluced emf ε
Pout = εIr = 98.2 V x 3.69 A = 362 W
Problem#3
A shunt-wound dc motor with the field coils and rotor connected in parallel (see Fig.1) operates from a 120-V dc power line. The resistance of the field windings, Rf is 218 Ω. The resistance of the rotor, Rr is 5.9 Ω. When the motor is running, the rotor develops an emf ε. The motor draws a current of 4.82 A from the line. Friction losses amount to 45.0 W. Compute (a) the field current; (b) the rotor current; (c) the emf (d) the rate of development of thermal energy in the field windings; (e) the rate of development of thermal energy in the rotor; (f) the power input to the motor; (g) the efficiency of the motor.
Answer:
Known:
Potential difference, Vab = 120 V
The motor draws a current of I = 4.82 A
Resistance of the field windings, Rf = 218 Ω
Resistance of the rotor, Rr = 5.9 Ω
(a) the field current is
If = Vab/Rf = 120 V/218 Ω = 0.550 A
(b) the rotor current;
Ir = Itotal – If = 4.82 A – 0.550 A = 4.27 A
(c) the emf is
V = ε + IrRr or
ε = V – IR = 120 V – 4.27 A x 5.9 Ω = 94.8 V
(d) the rate of development of thermal energy in the field windings is
Pf = If2Rf = (0.550 A)2(218 Ω) = 65.9 W
(e) the rate of development of thermal energy in the rotor is
Pr = Ir2Rr = (4.27 A)2(5.9 Ω) = 108 W
(f) the power input to the motor is
Pin = VI = 120 V x 4.82 A = 578 W
(g) the efficiency of the motor is
Efficiency = Pout/Pin
With, Pout = Pin – Pf – Pr – Ploss = 578 W – 65.9 W – 108 W – 45.0 W = 359 W
Then,
Efficiency = 359 W/578 W = 0.621
Vf = IfR
If = Vf/R = 120 V/106 Ω = 1.13 A
(b) the current in the rotor is
I – If – Ir = 0 (applying the juction rule to point a in the circuit diagram)
Ir = I – If = 4.82 A – 1.13 A = 3.69 A
(c) the induced emf developed by the motor is
ε = V – IrRr = 120 V – 3.69 A x 5.9 Ω = 98.2 V
(d) The mechanical power output is the electrical power input minus the rate of dissipation of electrical energy in the resistance of the motor:
Electrical power input to the rotor
Pin = IV = (4.82 A)(120 V) = 578 W
Electrical power loss in the to resistances
Ploss = If2Rf + Ir2R
Ploss = (1.13 A)2(106 Ω) + (3.69 A)2(5.9 Ω) = 216 W
Mechanical power output
Pout = Pin – Ploss = 578 W – 216 W = 362 W
Or the mechanical power output is the power assosiated with the incluced emf ε
Pout = εIr = 98.2 V x 3.69 A = 362 W
Problem#3
A shunt-wound dc motor with the field coils and rotor connected in parallel (see Fig.1) operates from a 120-V dc power line. The resistance of the field windings, Rf is 218 Ω. The resistance of the rotor, Rr is 5.9 Ω. When the motor is running, the rotor develops an emf ε. The motor draws a current of 4.82 A from the line. Friction losses amount to 45.0 W. Compute (a) the field current; (b) the rotor current; (c) the emf (d) the rate of development of thermal energy in the field windings; (e) the rate of development of thermal energy in the rotor; (f) the power input to the motor; (g) the efficiency of the motor.
Answer:
Known:
Potential difference, Vab = 120 V
The motor draws a current of I = 4.82 A
Resistance of the field windings, Rf = 218 Ω
Resistance of the rotor, Rr = 5.9 Ω
(a) the field current is
If = Vab/Rf = 120 V/218 Ω = 0.550 A
(b) the rotor current;
Ir = Itotal – If = 4.82 A – 0.550 A = 4.27 A
(c) the emf is
V = ε + IrRr or
ε = V – IR = 120 V – 4.27 A x 5.9 Ω = 94.8 V
(d) the rate of development of thermal energy in the field windings is
Pf = If2Rf = (0.550 A)2(218 Ω) = 65.9 W
(e) the rate of development of thermal energy in the rotor is
Pr = Ir2Rr = (4.27 A)2(5.9 Ω) = 108 W
(f) the power input to the motor is
Pin = VI = 120 V x 4.82 A = 578 W
(g) the efficiency of the motor is
Efficiency = Pout/Pin
With, Pout = Pin – Pf – Pr – Ploss = 578 W – 65.9 W – 108 W – 45.0 W = 359 W
Then,
Efficiency = 359 W/578 W = 0.621
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