Question

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

• A. $E=20+2L$
• B. $E=80+2L$
• C. $E=20+8L$
• D. $E=80+8L$

Answer 1

The correct option is A $E=20+2L$

The power source characteristic can be written analytically as

$E=80-\frac{80}{800}I.......\left(a\right)$

$\text{The arc characterisitc is given as}$

$I=aL+b$

$\text{Where}a\text{and}b\text{are constnat}$

$\text{Given data :}$

$I=500A$

$L=5.00mm$

$\therefore 500=5a+b\text{.........(i)}$

$I=460A$

$L=7.00mm$

$\therefore 460=7a+b\text{........ (ii)}$

$\text{Solving Eqs. (i) and (ii), we get}$

$a=-20$

$b=600$

$\text{Then arc characteristic equation}$

$I=-20L+600\text{.........(b)}$

$\text{From equations (a) and (b)}$

$E=80-\frac{80}{800}(-20L+600)$

$=80-0.1(-20L+600)$

$=80+2L-60$

$E=20+2L$