GATE 2022 ME (ME2) – Question 57
In a direct current arc welding process, the power source has an open circuit voltage of 100 V and short circuit current of 1000 A. Assume a linear relationship between voltage and current. The arc voltage (V) varies with the arc length (l) as $V=10+5l$, where V is in volts and l is in mm. The maximum available arc power during the process is ______ kVA (in integer).
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Correct answer: 25
Explanation
**Step 1: power source characteristic.** Open circuit voltage 100 V and short circuit current 1000 A, linear in between:
$$V=100-\frac{100}{1000}\,I=100-0.1\,I .$$
**Step 2: power delivered.**
$$P=VI=100I-0.1I^2 .$$
**Step 3: maximise.**
$$\frac{dP}{dI}=100-0.2I=0\;\Rightarrow\;I=500\text{ A},\qquad V=100-0.1(500)=50\text{ V}.$$
$$P_{max}=50\times500=25\,000\text{ VA}=\mathbf{25\ kVA}.$$
(The arc characteristic $V=10+5l$ gives $50=10+5l$, so $l=8$ mm, a feasible arc length.)