Question

In a DC arc welding operation, the voltage-arc length characteristic was obtained as \(V_{\text{arc}} = 20 + 5L\), where the arc length L was varied between 5 mm and 7mm. Here \(V_{\text{arc}}\) denotes the arc voltage in volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristics, the open circuit voltage and short circuit current for the welding operation are:

• A. 45 V, 450 A
• B. 75 V, 550 A
• C. 95 V, 950 A
• D. 150 V, 1500 A

Hint:

Always associate the minimum arc length (lower voltage) with the maximum current, and the maximum arc length (higher voltage) with the minimum current.
This inverse pairing is a fundamental characteristic of welding arcs.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to determine the open-circuit voltage (\(V_{\text{oc}}\)) and short-circuit current (\(I_{\text{sc}}\)) of a welding power source with a linear characteristic.
We are given the relationship between arc voltage and arc length, as well as the ranges for the arc length and current.

Step 2: Key Formula or Approach:
A linear power source characteristic is given by the equation:
\[ \frac{V}{V_{\text{oc}}} + \frac{I}{I_{\text{sc}}} = 1 \implies V = V_{\text{oc}} - \left(\frac{V_{\text{oc}}}{I_{\text{sc}}}\right) I \]
This can be rewritten in the form:
\[ V = V_{\text{oc}} - k I \]
where \(k\) is a constant.

Step 3: Detailed Explanation:

•
Step 3.1: Find the voltage limits:
For the minimum arc length, \(L_1 = 5\text{ mm}\):
\[ V_1 = 20 + 5(5) = 45\text{ V} \]
For the maximum arc length, \(L_2 = 7\text{ mm}\):
\[ V_2 = 20 + 5(7) = 55\text{ V} \]

•
Step 3.2: Pair the voltages with corresponding currents:
In welding, a longer arc (higher voltage) corresponds to a lower current, and a shorter arc (lower voltage) corresponds to a higher current.
Therefore:
At \(V_1 = 45\text{ V}\), the current is \(I_1 = 500\text{ A}\).
At \(V_2 = 55\text{ V}\), the current is \(I_2 = 400\text{ A}\).

•
Step 3.3: Set up and solve the linear system:
Using the linear characteristic equation \(V = V_{\text{oc}} - k I\):
\[ 45 = V_{\text{oc}} - 500 k \quad \text{--- (Equation 1)} \]
\[ 55 = V_{\text{oc}} - 400 k \quad \text{--- (Equation 2)} \]
Subtract Equation 1 from Equation 2:
\[ 10 = 100 k \implies k = 0.1\text{ V/A} \]
Substitute \(k = 0.1\) back into Equation 2:
\[ 55 = V_{\text{oc}} - 400(0.1) \]
\[ 55 = V_{\text{oc}} - 40 \implies V_{\text{oc}} = 95\text{ V} \]

•
Step 3.4: Calculate the short-circuit current (\(I_{\text{sc}}\)):
When the system is short-circuited, the voltage is zero (\(V = 0\)):
\[ 0 = 95 - 0.1 I_{\text{sc}} \implies 0.1 I_{\text{sc}} = 95 \implies I_{\text{sc}} = 950\text{ A} \]

Step 4: Final Answer:
The open circuit voltage is \(95\text{ V}\) and the short circuit current is \(950\text{ A}\).