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A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.
The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)
The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)
[ME 2007]
- a)E = 20 + 2L
- b)E = 20 + 8L
- c)E = 80 + 2L
- d)E = 80 + 8L
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A direct current welding machine with a linear power source characteri...
The power source characteristic can be written analytically as
The arc characteristic is given as
I = aL + b
where a and b are constant
Given,
When I = 500A
then L = 5.00 mm
∴ 500 = 5a + b ...(i)
when I = 460A
then L = 7.00 mm
∴ 460 = 7a + b ...(ii)
Solving Eqs. (i) and (ii, we get
a = –20
and b = 600
Then arc characteristic equation
I = –20L + 600
From equations (a) and (b)
= 80 – 0.1(–20L + 600)
= 80 + 2L – 60
= 20 + 2L
Most Upvoted Answer
A direct current welding machine with a linear power source characteri...
The power source characteristic can be written analytically as
The arc characteristic is given as
I = aL + b
where a and b are constant
Given,
When I = 500A
then L = 5.00 mm
∴ 500 = 5a + b ...(i)
when I = 460A
then L = 7.00 mm
∴ 460 = 7a + b ...(ii)
Solving Eqs. (i) and (ii, we get
a = –20
and b = 600
Then arc characteristic equation
I = –20L + 600
From equations (a) and (b)
= 80 – 0.1(–20L + 600)
= 80 + 2L – 60
= 20 + 2L
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Community Answer
A direct current welding machine with a linear power source characteri...
Given Data:
Open circuit voltage (Voc) = 80 V
Short circuit current (Isc) = 800 A
Measured arc current (I) = 500 A (corresponding to arc length L = 5.0 mm)
Measured arc current (I) = 460 A (corresponding to arc length L = 7.0 mm)
Explanation:
To find the linear voltage-arc length characteristic, we need to determine the equation that relates voltage (E) and arc length (L).
Step 1: Calculate the voltage drop across the arc using the measured current and the short circuit current:
Voltage drop across the arc (Vdrop) = Voc - I = 80 V - 500 A = -420 V
Voltage drop across the arc (Vdrop) = Voc - I = 80 V - 460 A = -380 V
Step 2: Calculate the voltage difference between the two measured arc lengths:
Voltage difference (ΔV) = Vdrop (L1) - Vdrop (L2) = -420 V - (-380 V) = -40 V
Step 3: Calculate the difference in arc length:
Arc length difference (ΔL) = L1 - L2 = 5.0 mm - 7.0 mm = -2.0 mm
Step 4: Calculate the slope of the voltage-arc length characteristic:
Slope (m) = ΔV / ΔL = -40 V / -2.0 mm = 20 V/mm
Step 5: Write the linear equation:
E = m * L + C
where E is the voltage, L is the arc length, m is the slope, and C is the constant.
Step 6: Substitute the values of slope and one set of measured values (E = 80 V, L = 5.0 mm) to find the constant C:
80 V = (20 V/mm) * 5.0 mm + C
80 V = 100 V + C
C = -20 V
Step 7: Substitute the values of slope and constant into the equation:
E = 20 V/mm * L - 20 V
Conclusion:
The linear voltage-arc length characteristic of the welding arc can be given as:
E = 20 V/mm * L - 20 V, which is equivalent to option A: E = 20 * 2L.
Open circuit voltage (Voc) = 80 V
Short circuit current (Isc) = 800 A
Measured arc current (I) = 500 A (corresponding to arc length L = 5.0 mm)
Measured arc current (I) = 460 A (corresponding to arc length L = 7.0 mm)
Explanation:
To find the linear voltage-arc length characteristic, we need to determine the equation that relates voltage (E) and arc length (L).
Step 1: Calculate the voltage drop across the arc using the measured current and the short circuit current:
Voltage drop across the arc (Vdrop) = Voc - I = 80 V - 500 A = -420 V
Voltage drop across the arc (Vdrop) = Voc - I = 80 V - 460 A = -380 V
Step 2: Calculate the voltage difference between the two measured arc lengths:
Voltage difference (ΔV) = Vdrop (L1) - Vdrop (L2) = -420 V - (-380 V) = -40 V
Step 3: Calculate the difference in arc length:
Arc length difference (ΔL) = L1 - L2 = 5.0 mm - 7.0 mm = -2.0 mm
Step 4: Calculate the slope of the voltage-arc length characteristic:
Slope (m) = ΔV / ΔL = -40 V / -2.0 mm = 20 V/mm
Step 5: Write the linear equation:
E = m * L + C
where E is the voltage, L is the arc length, m is the slope, and C is the constant.
Step 6: Substitute the values of slope and one set of measured values (E = 80 V, L = 5.0 mm) to find the constant C:
80 V = (20 V/mm) * 5.0 mm + C
80 V = 100 V + C
C = -20 V
Step 7: Substitute the values of slope and constant into the equation:
E = 20 V/mm * L - 20 V
Conclusion:
The linear voltage-arc length characteristic of the welding arc can be given as:
E = 20 V/mm * L - 20 V, which is equivalent to option A: E = 20 * 2L.
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A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer?
Question Description
A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer?.
A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer?.
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A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A direct current welding machine with a linear power source characteristic provides open circuit voltage of 80 V and short circuit current of 800 A. During welding with the machine, the measured arc current is 500 A corresponding to an arc length of 5.0 mm and the measured arc current is 460 A corresponding to an arc length of 7.0 mm.The linear voltage (E) - arc length (L) characteristic of the welding arc can be given as (where E is in Voit and L is in mm)[ME 2007]a)E = 20 + 2Lb)E = 20 + 8Lc)E = 80 + 2Ld)E = 80 + 8LCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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