Question

The voltage arc length characteristics of a $DC$ arc is given by $V=20+40L$, where $L=arc$ length in $cm$. The power source characteristics can be approximated by a straight line. Open circuit is $80V$ and short circuit current $1000Amps$. The optimum arc length in $mm$

• A. $1$
• B. $0.5$
• C. $10$
• D. $5$

Answer 1

The correct option is D $5$

$\text{Voltage arc characteristic}$

$V=20+40L\text{....... (i)}$

$\text{Where,}L=arc\text{length}\left(cm\right)$

$V_{OC}=80V$

$I_{SC}=1000Amps$

$\text{Current characteristics}$

$V=V_{OC}-\frac{V_{OC}}{I_{SC}}I$

$V=80-\frac{80}{1000}I\text{.....(ii)}$

$\text{By equations (i) and (ii)}$

$20+40L=80-\frac{80}{1000}I$

$\frac{80}{1000}I=80-20-40L$

$\frac{8}{1000}I=60-40L$

$\text{By power source characteristics}$

$P=VI$

$P=(20+40L)(60-40L)\frac{100}{8}I$

$P=(1200-800L+2400L-1600L^2)\frac{100}{8}I$

$\text{For optimum arc length.}$

$\frac{dP}{dL}=0$

$1600-1600\times 2L=0$

$L=0.5cm=5mm$