Question

$\text{The power source characteristics is given by}$
$I^2=-350(V-90)$
$\text{and the voltage-length characteristics of a DC arc is expressed by}$ $V=25+5l$,
$\text{where}{}^{\prime }{l}^{\prime }\text{is the arc length in mm, and}{}^{\prime }{V}^{\prime }\text{is the arc voltage.}$
$\text{Then change in welding current (Amp) for a change in are length}$
$\text{from 5 mm to 7 mm is\_\_\_\_\_\_\_\_}$

• A. 10 Amp
• B. 15.85 Amp
• C. 20 Amp
• D. 14.58 Amp

Answer 1

The correct option is B 15.85 Amp

The power source charactristics is

$I^2=-350(V-90)$.............(1)

The voltage-length characteristics of arc is

$V=25+5l$...................(2)

$Forl=5\text{mm}$

Using equestions (2)

$V=25+5\times 5=50$ volts

Using equation (1)

$I^2=-350(50-90)$

$=350\times 40$

$\therefore I=118.32\text{Amp}$

$For\ell =7\text{mm}$

$V=25+5\times 7=60\text{volts}$

$And{I}^2=-350(60-90)=350\times 30$

$\therefore I=102.47\text{Amp}$

Hence, change in wedding current is

$\text{(118.32-102.47) Amp = 15.85 Amp}$