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Two inductors L1 = 20 mH and L2 = 40 mH are connected in series so that their equivalent inductance is 50 mH. The mutual inductance between the two coils is _______.
- a)15 mH
- b)10 mH
- c)20 mH
- d)5 mH
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Two inductors L1 = 20 mH and L2 = 40 mH are connected in series so tha...
To find the mutual inductance between the two coils, we can use the formula:
Lm = √(L1 * L2 - Leq^2)
where Lm is the mutual inductance, L1 and L2 are the individual inductances, and Leq is the equivalent inductance.
Given:
L1 = 20 mH
L2 = 40 mH
Leq = 50 mH
Substituting the given values into the formula, we get:
Lm = √(20 mH * 40 mH - 50 mH^2)
= √(800 mH^2 - 2500 mH^2)
= √(-1700 mH^2)
Since the square root of a negative number is not defined in the real number system, the mutual inductance cannot be determined based on the given values. Therefore, none of the answer options provided (a, b, c, or d) is correct.
It's possible that there may be a mistake in the given information or the question itself. Please double-check the values or provide any additional information if available to obtain the correct answer.
Lm = √(L1 * L2 - Leq^2)
where Lm is the mutual inductance, L1 and L2 are the individual inductances, and Leq is the equivalent inductance.
Given:
L1 = 20 mH
L2 = 40 mH
Leq = 50 mH
Substituting the given values into the formula, we get:
Lm = √(20 mH * 40 mH - 50 mH^2)
= √(800 mH^2 - 2500 mH^2)
= √(-1700 mH^2)
Since the square root of a negative number is not defined in the real number system, the mutual inductance cannot be determined based on the given values. Therefore, none of the answer options provided (a, b, c, or d) is correct.
It's possible that there may be a mistake in the given information or the question itself. Please double-check the values or provide any additional information if available to obtain the correct answer.
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Community Answer
Two inductors L1 = 20 mH and L2 = 40 mH are connected in series so tha...
Concept:
The equivalent inductance of series aiding connection is
Leq = L1 + L2 + 2M
The equivalent inductance of series opposing connection is
Leq = L1 + L2 – 2M
Calculation:
Given:
L1 = 20 mH
L2 = 40 mH
Leq = 50 mH
From the given data we can conclude that series opposing connection will be considered as the mutual index cannot be negative.
Leq = L1 + L2 – 2M
50 = 40 + 20 - 2M
M = 5 mH
The equivalent inductance of series aiding connection is
Leq = L1 + L2 + 2M
The equivalent inductance of series opposing connection is
Leq = L1 + L2 – 2M
Calculation:
Given:
L1 = 20 mH
L2 = 40 mH
Leq = 50 mH
From the given data we can conclude that series opposing connection will be considered as the mutual index cannot be negative.
Leq = L1 + L2 – 2M
50 = 40 + 20 - 2M
M = 5 mH
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Two inductors L1 = 20 mH and L2 = 40 mH are connected in series so that their equivalent inductance is 50 mH. The mutual inductance between the two coils is _______.a)15 mHb)10 mHc)20 mHd)5 mHCorrect answer is option 'D'. Can you explain this answer?
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