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A DC 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 Amperes) should be set as _____.
(Important - Enter only the numerical value in the answer)
Correct answer is '150'. Can you explain this answer?
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A DC welding power source has a linear voltage-current (V-I) character...
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A DC welding power source has a linear voltage-current (V-I) character...
Linear Voltage-Current Characteristic
The linear voltage-current (V-I) characteristic of a DC welding power source describes how the voltage across the terminals of the power source changes with the current flowing through it. In this case, the power source has an open circuit voltage of 80V and a short circuit current of 300A.
Maximum Arc Power
To achieve maximum arc power, the current needs to be set at a specific value. This value can be determined by considering the power equation, which states that power (P) is equal to the product of voltage (V) and current (I), i.e., P = V * I.
Calculating Maximum Power
To calculate the maximum power, we need to find the value of current that maximizes the power output. This can be done by considering the V-I characteristic.
- At open circuit (when no current is flowing), the voltage is 80V.
- At short circuit (when the current is maximum), the current is 300A.
To find the maximum power, we need to find the point on the V-I characteristic curve that lies on the line connecting the open circuit voltage and the short circuit current.
Linear Interpolation
To find this point, we can use linear interpolation. Linear interpolation involves determining the value of a variable based on its position relative to two known values.
In this case, we want to find the current value that corresponds to a voltage of 80V when the current is 300A.
Using linear interpolation, we can calculate the current as follows:
(V1 - V) / (V1 - V2) = (I1 - I) / (I1 - I2)
Where:
- V1 = Open circuit voltage = 80V
- V2 = Short circuit voltage = 0V (since it is a short circuit)
- I1 = Short circuit current = 300A
- I2 = Open circuit current = 0A (since it is an open circuit)
- V = Voltage = 80V (the desired value)
- I = Current (the desired value)
Substituting the known values into the equation, we get:
(80 - 80) / (80 - 0) = (300 - I) / (300 - 0)
0 / 80 = (300 - I) / 300
0 = (300 - I) / 300
0 = 300 - I
I = 300
Therefore, the current should be set at 300 Amperes for maximum arc power.
The linear voltage-current (V-I) characteristic of a DC welding power source describes how the voltage across the terminals of the power source changes with the current flowing through it. In this case, the power source has an open circuit voltage of 80V and a short circuit current of 300A.
Maximum Arc Power
To achieve maximum arc power, the current needs to be set at a specific value. This value can be determined by considering the power equation, which states that power (P) is equal to the product of voltage (V) and current (I), i.e., P = V * I.
Calculating Maximum Power
To calculate the maximum power, we need to find the value of current that maximizes the power output. This can be done by considering the V-I characteristic.
- At open circuit (when no current is flowing), the voltage is 80V.
- At short circuit (when the current is maximum), the current is 300A.
To find the maximum power, we need to find the point on the V-I characteristic curve that lies on the line connecting the open circuit voltage and the short circuit current.
Linear Interpolation
To find this point, we can use linear interpolation. Linear interpolation involves determining the value of a variable based on its position relative to two known values.
In this case, we want to find the current value that corresponds to a voltage of 80V when the current is 300A.
Using linear interpolation, we can calculate the current as follows:
(V1 - V) / (V1 - V2) = (I1 - I) / (I1 - I2)
Where:
- V1 = Open circuit voltage = 80V
- V2 = Short circuit voltage = 0V (since it is a short circuit)
- I1 = Short circuit current = 300A
- I2 = Open circuit current = 0A (since it is an open circuit)
- V = Voltage = 80V (the desired value)
- I = Current (the desired value)
Substituting the known values into the equation, we get:
(80 - 80) / (80 - 0) = (300 - I) / (300 - 0)
0 / 80 = (300 - I) / 300
0 = (300 - I) / 300
0 = 300 - I
I = 300
Therefore, the current should be set at 300 Amperes for maximum arc power.
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A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer?
Question Description
A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer?.
A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer?.
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A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer?, a detailed solution for A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? has been provided alongside types of A DC welding power source has a linear voltage-current (V-I) characteristic with open circuitvoltage of 80 V and a short circuit current of 300 A. For maximum arc power, the current (inAmperes) should be set as _____.(Important - Enter only the numerical value in the answer)Correct answer is '150'. Can you explain this answer? theory, EduRev gives you an
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