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An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where i is
- a)14 A
- b)about 20 A
- c)7 A
- d)about 10 A
Correct answer is option 'D'. Can you explain this answer?
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An alternating current having peak value 14 A is used to heat a metal ...
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An alternating current having peak value 14 A is used to heat a metal ...
Explanation:
Peak value of AC current:
- The peak value of the alternating current (AC) is given as 14 A.
Heating effect of current:
- The heating effect of current is directly proportional to the square of the current flowing through the wire (P ∝ I²).
Constant current for same heating effect:
- To produce the same heating effect as the AC current with a peak value of 14 A, we need to find a constant current (i) that would result in the same power dissipation.
- Since power is proportional to the square of the current, we can equate the power dissipation for the AC current (14 A) to the power dissipation for the constant current (i).
Calculation:
- Peak value of AC current (I_AC) = 14 A
- Constant current (i) for the same heating effect = i
- Equating the power dissipation for both currents:
I_AC² = i²
(14 A)² = i²
196 A² = i²
i ≈ 14 A
Conclusion:
- Therefore, to produce the same heating effect as the alternating current with a peak value of 14 A, a constant current of about 14 A (approximately 10 A) can be used.
Peak value of AC current:
- The peak value of the alternating current (AC) is given as 14 A.
Heating effect of current:
- The heating effect of current is directly proportional to the square of the current flowing through the wire (P ∝ I²).
Constant current for same heating effect:
- To produce the same heating effect as the AC current with a peak value of 14 A, we need to find a constant current (i) that would result in the same power dissipation.
- Since power is proportional to the square of the current, we can equate the power dissipation for the AC current (14 A) to the power dissipation for the constant current (i).
Calculation:
- Peak value of AC current (I_AC) = 14 A
- Constant current (i) for the same heating effect = i
- Equating the power dissipation for both currents:
I_AC² = i²
(14 A)² = i²
196 A² = i²
i ≈ 14 A
Conclusion:
- Therefore, to produce the same heating effect as the alternating current with a peak value of 14 A, a constant current of about 14 A (approximately 10 A) can be used.
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An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where i isa)14 Ab)about 20 Ac)7 Ad)about 10 ACorrect answer is option 'D'. Can you explain this answer?
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