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Question

In gas metal arc welding, the power source characteristics is
$V_{p} = 36 - \frac{I}{60}$
and the arc characteristics is
$V_{a} = 2 l_{a} + 27$ .
Where ' $V_{p}$ ' and ' $V_{a}$ ' are voltage, 'I' is current and l $_{a}$ is arc length in mm. The change in power (VA) of the arc, if the arc length is changed from 3 mm to 4 mm is

A

3840

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B

5940

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C

2100

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D

7200

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Solution

The correct option is A 3840
Power source characteristics is

$V_{p} = 36 - \frac{I}{60}$ ......(1)

$Arc characteristics (V_{a}) = 2 l_{a} + 27 ....(2)$

Equating these equations (1) & (2)

$36 - \frac{I}{60} = 2 l_{a} + 27$

I = $60 \times (9 - 2 l_{a})$

Power
$(P_{2}) = (2 l_{a} + 27) \times 60 \times (9 - 2 l_{a}) When l_{a} = 3 m m P_{3} = (2 \times 3 + 27) \times 60 \times (9 - 2 \times 3) = 33 \times 60 \times 3 = 5940 V A When l_{a} = 4 m m P_{4} = (2 \times 4 + 27) \times 60 \times (9 - 2 \times 4) = 35 \times 60 = 2100 V A ∴ Change in power of Arc$
= 5940 - 2100 = 3840 VA

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