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Standard XII

Physics

Transistor

ADC welding p...

Question

A $D C$ welding power source has a linear voltage-current $(V - I)$ characteristic with open circuit voltage of $80 V$ and a short circuit current of $300 A$ . For maximum arc power, the current (in Ampere) should be set as
150

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Solution

The correct option is A 150
$V = V_{O C} (\frac{V_{O C}}{I_{S C}}) I$

$V = 80 - (\frac{80}{300}) I$

$P = V I$

$= (80 - 0.2666 I) I$

$= 80 I - 0.26666 I^{2}$

$For power to be maximum$

$(\frac{d P}{d I}) = 0$

$\Rightarrow 80 - 2 \times 0.2666 I = 0$

$\Rightarrow I = 150 A$

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ADC welding power source has a linear voltage-current (V−I) characteristic with open circuit voltage of 80V and a short circuit current of 300A. For maximum arc power, the current (in Ampere) should be set as150

The correct option is A 150V=VOC(VOCISC)I V=80−(80300)I P=VI =(80−0.2666I)I =80I−0.26666I2 For power to be maximum (dPdI)=0 ⇒80−2×0.2666I=0 ⇒I=150A

A $D C$ welding power source has a linear voltage-current $(V - I)$ characteristic with open circuit voltage of $80 V$ and a short circuit current of $300 A$ . For maximum arc power, the current (in Ampere) should be set as
150

The correct option is A 150
$V = V_{O C} (\frac{V_{O C}}{I_{S C}}) I$

$V = 80 - (\frac{80}{300}) I$

$P = V I$

$= (80 - 0.2666 I) I$

$= 80 I - 0.26666 I^{2}$

$For power to be maximum$

$(\frac{d P}{d I}) = 0$

$\Rightarrow 80 - 2 \times 0.2666 I = 0$

$\Rightarrow I = 150 A$