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A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency?
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A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms ...
**Given Data:**
- Voltage (V) = 100V
- Power (P) = 1 kW
- Armature Resistance (Ra) = 1.25 ohms
- Speed (N) = 600 rpm
- Field Current at No Load (I0) = 1 A
- Field Resistance Reduction at Full Load = 50%
**Calculations:**
**i) Armature Current at Full Load:**
- Power (P) = V * Ia (where Ia is the Armature Current)
- 1 kW = 100V * Ia
- Ia = 1 kW / 100V
- Ia = 10 A
**ii) Torque and Speed at Full Load:**
- Power (P) = T * N (where T is the Torque and N is the Speed)
- 1 kW = T * 600 rpm
- T = 1 kW / 600 rpm
- T = 1.67 Nm
**iii) Electrical Efficiency:**
- Efficiency (η) = (Output Power / Input Power) * 100
- Output Power = Power (P) = 1 kW
- Input Power = V * Ia (where V is the Voltage and Ia is the Armature Current)
- Input Power = 100V * 10 A = 1000 W
- Efficiency (η) = (1 kW / 1000 W) * 100
- Efficiency (η) = 100%
**Explanation:**
**i) Armature Current at Full Load:**
- The power of the motor is given as 1 kW, and the voltage is given as 100V. Using the power equation, we can calculate the armature current Ia as 10 A.
**ii) Torque and Speed at Full Load:**
- The power equation is used again to find the torque T. By dividing the power by the speed, we find that the torque required at full load is 1.67 Nm.
**iii) Electrical Efficiency:**
- The electrical efficiency is calculated by dividing the output power by the input power and multiplying it by 100. In this case, the output power is 1 kW, and the input power is 1000 W (calculated using the voltage and armature current). Therefore, the electrical efficiency is 100%.
**In summary, at full load:**
- The armature current is 10 A.
- The torque is 1.67 Nm.
- The speed remains at 600 rpm.
- The electrical efficiency is 100%.
- Voltage (V) = 100V
- Power (P) = 1 kW
- Armature Resistance (Ra) = 1.25 ohms
- Speed (N) = 600 rpm
- Field Current at No Load (I0) = 1 A
- Field Resistance Reduction at Full Load = 50%
**Calculations:**
**i) Armature Current at Full Load:**
- Power (P) = V * Ia (where Ia is the Armature Current)
- 1 kW = 100V * Ia
- Ia = 1 kW / 100V
- Ia = 10 A
**ii) Torque and Speed at Full Load:**
- Power (P) = T * N (where T is the Torque and N is the Speed)
- 1 kW = T * 600 rpm
- T = 1 kW / 600 rpm
- T = 1.67 Nm
**iii) Electrical Efficiency:**
- Efficiency (η) = (Output Power / Input Power) * 100
- Output Power = Power (P) = 1 kW
- Input Power = V * Ia (where V is the Voltage and Ia is the Armature Current)
- Input Power = 100V * 10 A = 1000 W
- Efficiency (η) = (1 kW / 1000 W) * 100
- Efficiency (η) = 100%
**Explanation:**
**i) Armature Current at Full Load:**
- The power of the motor is given as 1 kW, and the voltage is given as 100V. Using the power equation, we can calculate the armature current Ia as 10 A.
**ii) Torque and Speed at Full Load:**
- The power equation is used again to find the torque T. By dividing the power by the speed, we find that the torque required at full load is 1.67 Nm.
**iii) Electrical Efficiency:**
- The electrical efficiency is calculated by dividing the output power by the input power and multiplying it by 100. In this case, the output power is 1 kW, and the input power is 1000 W (calculated using the voltage and armature current). Therefore, the electrical efficiency is 100%.
**In summary, at full load:**
- The armature current is 10 A.
- The torque is 1.67 Nm.
- The speed remains at 600 rpm.
- The electrical efficiency is 100%.
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A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency?
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A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency?.
A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100V , 1 Kw Dc shunt motor with an armature resistance of 1.25 ohms rotates at 600rpm. A variable field resistance draws a current of 1 A at no load. Given that field resistance is reduced by 50 % at full load. Neglecting the mechanical and iron losses. Calculate the full load (i) Armature current ii) Torque and Speed (iii) Electrical efficiency?.
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