A direct current welding machine with a linear power source

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A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.
The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)

1. E = 20 + 2L
2. E = 20 + 8L
3. E = 80 + 2L
4. E = 80 + 8L

Correct Option: A

The power source characteristic can be written analytically as

E = 80 −	80	I (a)
	800

The arc characteristic is given as I = aL + b
where a and b are constant Given,
When I = 500A then L = 5.00 mm
∴ 500 = 5a + b ...(i)
when I = 460A
then L = 7.00 mm
∴ 460 = 7a + b ...(ii)
Solving Eqs. (i) and (ii, we get a = –20
and b = 600
Then arc characteristic equation I = –20L + 600
From equations (a) and (b)

E = 80 −	80	(−20L + 600)
	800

= 80 – 0.1(–20L + 600)
= 80 + 2L – 60
= 20 + 2L