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Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

In a DC arc w...

Question

In a $D C$ arc welding operation, the voltage length characteristic was obtained as $V_{a r c} = 20 + 5 l$ where the arc length $l$ was varied between $5 m m$ and $7 m m$ . Here $V_{a r c}$ denotes the arc voltage Volts. The arc current was varied from $400 A$ and $599 A$ . Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation is

150 V, 1500 A

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75 V, 750 A

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45 V, 450 A

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95 V, 950 A

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Solution

The correct option is D $95 V, 950 A$
$V_{a r c} = 20 + 5 l$

$l = 5 m m, V_{a r c} = 45 V$

$l = 7 m m, V_{a r c} = 55 V$

$V = V_{0} - \frac{V_{0}}{I_{S}} I$

$V_{0} = Open circuit voltage$

$I_{S} = Short circuit current$

$45 = V_{0} - \frac{V_{0}}{I_{S}} \times 500$

$55 = V_{0} - \frac{V_{0}}{I_{S}} \times 400$

$\Rightarrow V_{0} = 95 volts$

$I_{S} = 950 amp$

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The correct option is D 95V,950AVarc=20+5l l=5mm,Varc=45V l=7mm,Varc=55V V=V0−V0ISI V0= Open circuit voltage IS= Short circuit current 45=V0−V0IS×500 55=V0−V0IS×400 ⇒V0=95volts IS=950amp