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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
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- 45 V, 450 A
- 75 V, 750 A
- 95 V, 950 A
- 150 V, 1500 A
Correct Option: C
Var = 20 + 5l
For l = 5 mm, Varc = 45 V
For l = 7 mm, Varc = 55V
| V = V0 − | I | |
| IS |
where Vo is open circuit voltage and IS is short circuit current.
| ∴ 45 = V0 − | × 500 | |
| IS |
| and 55 = V0 − | × 400 | |
| IS |
By solving, we get
V0 = 95 Volts
IS = 950 A
