Question

The voltage length characteristics of a D.C. arc is given by V=20+30L, where ‘V’ is the arc voltage and ‘L’ is the length of arc in cm. The power source chracteristics is approximated by a straight line with an open circuit voltage is 60V and short circuit current is 200Amp. Determine the optimum arc length and the corresponding arc power.
0.33 cm and 1.99 KVA
0.25 cm and 2 KVA
0.33 mm and 2.99 KVA
0.33 cm and 2.99 KVA

Answer 1

Open circuit voltage , OCV = 60V { means emf of battery is 60V}
short circuit current , SCC = 200 Amp
Also given relation between Voltage and arc length is V = 20 + 30L

We know,
V = ξ - IR
Here ξ = 60V , R = ξ/SCC = 60/200 = 0.3
so, V = 60 - 0.3i
⇒20 + 30L = 60 -0.3i
⇒30L = 40 - 0.3i
⇒i = 400/3 - 100L , relation between current and arc length

Now, power = Vi
P = (20 + 30L)( 400/3 - 100L)
= 800/3 + 4000L - 2000L - 3000L²
Now, differentiate with respect to L for optimum arc at maximum power
dP/dL = 2000 -6000L = 0 ⇒ L = 2000/6000 = 1/3 = 0.33 cm

Hence, optimum arc length = 0.33cm

Now, arc length = 1/3 cm = 1/300 m
I = 400/3 - 100L = 400/3 - 100/300 = 399/3 = 133 Amp
V = 20 + 30L = 20 + 30/300 = 20.1 V

Now, Power = Vi = 133 × 20.1 = 2673.3 = 2.6733KVA

Hence, option (D) is correct .