Question

The voltage length characteristic of a direct area in an arc welding process is $V=(100+40l$, where $l$) is the length of the arc in $mm$ and $V$ is arc voltage in volts. During a welding operation, the arc length varies between $1$ and $2mm$ and the welding current is in the range $200-250A$. Assuming a linear power source, the short circuit current is

• A. 425

Answer 1

The correct option is A 425
$V=100+40L$

$\text{When,}l=1,I=250$

$V=(100+40\times 1)=140$

$=V_o-\frac{V_o}{I_o}\times 250$

$l=2,I=200$

$V=(100+40\times 2)$

$180=V_0-\frac{V_0}{I_0}\times 200$

$V_0=140+\frac{4}{5}\times 250=340$

$I=\frac{V_0\times 5}{4}=\frac{340\times 5}{4}=425A$