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A 220 V, PC series motor has armature and field resistances of 0.26 0 and 0.20 0 respectively. It takes acument of 50 A from the supply while running a 3000mm If a diverter resistance of 0.2.0 is connected across the field coil of the motor Calculate the new steady state armature current. Assume the load torque remains constant.?
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A 220 V, PC series motor has armature and field resistances of 0.26 0 ...
Problem Statement

We are given a PC series motor with the following parameters:
- Armature resistance (Ra) = 0.26 Ω
- Field resistance (Rf) = 0.20 Ω
- Supply voltage (Vs) = 220 V
- Supply current (Is) = 50 A
- Load torque remains constant.

A diverter resistance of 0.20 Ω is connected across the field coil of the motor. We need to calculate the new steady-state armature current.

Solution

To solve this problem, we can use the principle of electrical circuit analysis. We will analyze the circuit and calculate the new steady-state armature current.

Step 1: Calculate the armature current without the diverter resistance.

Using Ohm's law, we can calculate the armature current (Ia) without the diverter resistance.
Ia = (Vs - Is * Ra) / Ra
Ia = (220 - 50 * 0.26) / 0.26
Ia = (220 - 13) / 0.26
Ia = 807.69 A

Step 2: Calculate the total resistance in the field circuit.

The total resistance in the field circuit is the sum of the field resistance (Rf) and the diverter resistance.
Rtotal = Rf + Rdiverter
Rtotal = 0.20 + 0.20
Rtotal = 0.40 Ω

Step 3: Calculate the new steady-state armature current.

Using Ohm's law, we can calculate the new steady-state armature current (Ia_new) with the diverter resistance.
Ia_new = (Vs - If * Rtotal) / Ra
where If is the field current.

We can calculate the field current using Ohm's law.
If = Vs / (Rf + Rdiverter)
If = 220 / (0.20 + 0.20)
If = 220 / 0.40
If = 550 A

Now, substituting the values in the equation for Ia_new:
Ia_new = (220 - 550 * 0.40) / 0.26
Ia_new = (220 - 220) / 0.26
Ia_new = 0 A

Conclusion

The new steady-state armature current (Ia_new) with the diverter resistance is 0 A. This means that the diverter resistance effectively reduces the field current to zero, resulting in no armature current.
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A 220 V, PC series motor has armature and field resistances of 0.26 0 and 0.20 0 respectively. It takes acument of 50 A from the supply while running a 3000mm If a diverter resistance of 0.2.0 is connected across the field coil of the motor Calculate the new steady state armature current. Assume the load torque remains constant.?
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A 220 V, PC series motor has armature and field resistances of 0.26 0 and 0.20 0 respectively. It takes acument of 50 A from the supply while running a 3000mm If a diverter resistance of 0.2.0 is connected across the field coil of the motor Calculate the new steady state armature current. Assume the load torque remains constant.? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 220 V, PC series motor has armature and field resistances of 0.26 0 and 0.20 0 respectively. It takes acument of 50 A from the supply while running a 3000mm If a diverter resistance of 0.2.0 is connected across the field coil of the motor Calculate the new steady state armature current. Assume the load torque remains constant.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220 V, PC series motor has armature and field resistances of 0.26 0 and 0.20 0 respectively. It takes acument of 50 A from the supply while running a 3000mm If a diverter resistance of 0.2.0 is connected across the field coil of the motor Calculate the new steady state armature current. Assume the load torque remains constant.?.
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