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A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.
a) How many stages of starting resistance will be required to limit the current to the range specified?
b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.?
a) How many stages of starting resistance will be required to limit the current to the range specified?
b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.?
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A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.05...
Determining Stages of Starting Resistance
To limit the starting current to twice the rated value (i.e., 700 A for a 350 A rated motor), we first need to calculate the maximum allowable starting current.
Next, we can find the total resistance needed to limit this current using Ohm's Law (V = I × R).
Now, to determine the number of stages of resistance needed:
To limit the current effectively, a common approach is to have 3-5 stages. For example, if n = 4:
Voltage Cut-out for Each Stage
The voltage cut-out for each stage can be calculated at various armature currents until the rated current is reached.
This method ensures a smooth transition to full-load conditions, preventing excessive starting current.
To limit the starting current to twice the rated value (i.e., 700 A for a 350 A rated motor), we first need to calculate the maximum allowable starting current.
- Rated Current: 350 A
- Maximum Starting Current: 2 x 350 A = 700 A
Next, we can find the total resistance needed to limit this current using Ohm's Law (V = I × R).
- Voltage (V) = 250 V
- Armature Resistance (Ra) = 0.052 Ω
- Required Total Resistance (Rt) = V / I = 250 V / 700 A = 0.357 Ω
- Effective Resistance Needed = Rt - Ra = 0.357 Ω - 0.052 Ω = 0.305 Ω
Now, to determine the number of stages of resistance needed:
- Assuming equal segments, let the resistance of each segment be Rs.
- Let n be the number of stages.
- n × Rs = 0.305 Ω
To limit the current effectively, a common approach is to have 3-5 stages. For example, if n = 4:
- Rs = 0.305 Ω / 4 = 0.07625 Ω (approx. 0.076 Ω per segment)
Voltage Cut-out for Each Stage
The voltage cut-out for each stage can be calculated at various armature currents until the rated current is reached.
- Initial Current: 700 A
- Subsequent Current: 600 A, 500 A, etc., until 350 A.
- Voltage at cut-out stages can be calculated as: V = I × (Ra + n × Rs).
This method ensures a smooth transition to full-load conditions, preventing excessive starting current.
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A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.?
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A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.?.
A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100-hp, 250v, 350A shunt dc motor has an armature resistance of 0.052. It is desired to design a starter circuit for this motor which will limit the maximum starting current to twice its rated value and which will switch out sections of resistances as the armature current falls to its rated value.a) How many stages of starting resistance will be required to limit the current to the range specified?b) What must be the value of each segment of the resistor to be? At what voltage should each stage of the starting resistance be cut out.?.
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