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Standard X

Physics

Electric Current(I)

The arc lengt...

Question

The arc length voltage characteristic of a DC arc is given by the equation V = 24 + 4L where V is the arc voltage and 'L' is the arc length in mm. The static volt ampere characteristic of the power source is approximated by a straight line with no load voltage of 80 volts and the short circuit current of 600 amperes.

The power utilized is

120 W

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12 MW

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12 kW

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120 kW

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Solution

The correct option is C 12 kW
$V = 24 + 4 L O C V = 80 V S S C = 600 A \frac{V}{O C V} + \frac{I}{S S C} = 1 ∴ \frac{V}{80} + \frac{I}{600} = 1 P = V I = 600 V (1 - \frac{V}{80}) P = 600 (V - \frac{V^{2}}{80}) {For P =P}_{m a x} \frac{d P}{d V} = 0 ∴ 600 (1 - \frac{2 V}{80}) = 0 \Rightarrow V = 40 V o l t s And, I = 600 (1 - \frac{V}{80}) = 600 (1 - \frac{40}{80}) I = 300 A P = V I = 40 \times 300 = 12 k W$

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Similar questions

Q. The arc length voltage characteristic of a DC arc is given by the equation V = 24 + 4L where V is the arc voltage and 'L' is the arc length in mm. The static volt ampere characteristic of the power source is approximated by a straight line with no load voltage of 80 volts and the short circuit current of 600 amperes.

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The correct option is C 12 kWV=24+4LOCV=80VSSC=600AVOCV+ISSC=1∴V80+I600=1P=VI=600V(1−V80)P=600(V−V280)For P =PmaxdPdV=0∴600(1−2V80)=0⇒V=40VoltsAnd, I=600(1−V80)=600(1−4080)I=300AP=VI=40×300=12kW