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In a DC arc welding operation, the voltage-arc length characteristic was obtained as Varc = 20 + 5l where the arc length l was varied between 5 mm and 7 mm. Here Varc denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are
[ME 2012]
  • a)
    45 V, 450 A
  • b)
    75 V, 750 A
  • c)
    95 V, 950 A
  • d)
    150 V, 1500 A
Correct answer is option 'C'. Can you explain this answer?
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In a DC arc welding operation, the voltage-arc length characteristic w...

where Vo is open circuit voltage and IS is short circuit current.
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In a DC arc welding operation, the voltage-arc length characteristic w...
Given:
- Voltage-arc length characteristic: Varc = 20 + 5l
- Arc length, l, varies between 5 mm and 7 mm
- Arc current, I, varies from 400 A to 500 A

To find:
- Open circuit voltage, Voc
- Short circuit current, Isc

Solution:
1. Voltage-Arc Length Characteristic:
The given voltage-arc length characteristic equation is Varc = 20 + 5l, where Varc is the arc voltage in volts and l is the arc length in mm.

2. Variations in Arc Length and Arc Current:
The arc length, l, varies between 5 mm and 7 mm, and the arc current, I, varies from 400 A to 500 A.

3. Open Circuit Voltage:
The open circuit voltage, Voc, is the voltage across the welding machine when no current is flowing through the circuit. It can be obtained by substituting the minimum arc length value (l = 5 mm) into the voltage-arc length characteristic equation.

Varc = 20 + 5l
Voc = 20 + 5(5)
Voc = 20 + 25
Voc = 45 V

Therefore, the open circuit voltage for the welding operation is 45 V.

4. Short Circuit Current:
The short circuit current, Isc, is the current flowing through the circuit when the arc length is zero (l = 0). Since the given arc length variation starts from 5 mm, we need to find the short circuit current for an arc length of 5 mm.

Varc = 20 + 5l
400 = 20 + 5(5)
400 = 20 + 25
400 = 45Isc
Isc = 400/45
Isc = 8.89 A

Therefore, the short circuit current for the welding operation is 8.89 A.

Conclusion:
The open circuit voltage for the welding operation is 45 V, and the short circuit current is 8.89 A. Hence, the correct answer is option C (95 V, 950 A).
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In a DC arc welding operation, the voltage-arc length characteristic was obtained as Varc = 20 + 5l where the arc length l was varied between 5 mm and 7 mm. Here Varc denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are[ME 2012]a)45 V, 450 Ab)75 V, 750 Ac)95 V, 950 Ad)150 V, 1500 ACorrect answer is option 'C'. Can you explain this answer?
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