Respuesta :
Answer:
a) [tex]E_{b} = 221 V[/tex]
b) P = 13,800 W
c) [tex]P_{mech} = 13260 W[/tex]
di) [tex]I_{initial} = 1533.33 A[/tex]
dii) [tex]R_{start} = 1.85 ohms[/tex]
Explanation:
Voltage, Vt = 230 V
Armature current, [tex]I_{a} = 60 A[/tex]
Armature Resistance, [tex]R_{a} = 0.15 ohms[/tex]
a) The back emf is calculated as follows:
[tex]E_{b} = V_{t} - I_{a} R_{a} \\E_{b} = 230 - (60 * 0.15)\\E_{b} = 230 - 9\\E_{b} = 221 V[/tex]
b) The power supplied to the armature (W)
[tex]P =V_{t} I_{a}[/tex]
P = 230 * 60
P = 13,800 W
c) Mechanical power developed by the motor
[tex]P_{mech}[/tex] = Power supplied to the armature - Power lost in the armature
Power lost in the armature, [tex]P_{a} = I_{a} ^{2} R_{a}[/tex]
[tex]P_{a} = 60^{2} *0.15\\P_{a} = 540 W[/tex]
[tex]P_{mech} = 13800 - 540\\P_{mech} = 13260 W[/tex]
d)( i)The initial starting current if the motor is directly connected across the 230 V line
At starting, there is no back emf, [tex]E_{b} = 0[/tex]
[tex]V_{t} = I_{initial} R_{a}[/tex]
[tex]230 = I_{initial} * 0.15\\ I_{initial} = 230/0.15\\ I_{initial} = 1533.33 A[/tex]
ii) Value of the starting resistor needed to limit the initial current to 115 A
[tex]V_{t} = I_{initial} (R_{a} + R_{start})[/tex]
[tex]230= 115 (0.15 + R_{start})\\230/115 = 0.15 + R_{start}\\2 - 0.15 = R_{start}\\ R_{start} = 1.85 ohms[/tex]
