Question

A DC welding power source has a linear voltage-Current (V-I) characteristic with an open-circuit voltage of 80 Volt and a short circuit current of 300 A. For maximum arc power, the Current (in Amperes) should be set as

1. 1200 A
2. 150 A
3. 1500 A
4. 120 A

Answer 1

Correct Answer - Option 2 : 150 A

Concept:

The voltage supplied for welding is given by

\(V = {V_o} - \frac{{{V_o}}}{{{I_s}}}I\),

where V_o = Open circuit voltage, I_s = Short circuit current.

Power supplied is given by

\(P = VI = \left( {{V_o} - \frac{{{V_o}}}{{{I_s}}}I} \right) \times I\)

\(P = {V_o} \times I - \frac{{{V_o}}}{{{I_s}}}{I^2}\)

Now, For maximum arc power, \(\frac{{dP}}{{dI}} = 0\)

\(i.e.\;\frac{{dP}}{{dI}} = {V_o} - \frac{{2I{V_o}}}{{{I_s}}} = 0\)

∴ \({\bf{I}} = \frac{{{{\bf{I}}_{\bf{s}}}}}{2}\)

Calculation:

Given:

V_o = 80 V, I_s = 300 Ampere, I = ?

Now, we know that

\({\bf{I}} = \frac{{{{\bf{I}}_{\bf{s}}}}}{2} = \frac{{300}}{2}\)

∴ I = 150 Ampere