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Two coils with self-inductance of 100 mH and 200 mH and mutual inductance of 50 mH are connected in parallel aiding order. What will be the total inductance of the combination?
- a)250 mH
- b)4
- c)75 mH3.175 mH
- d)87.5 mH
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Two coils with self-inductance of 100 mH and 200 mH and mutual inducta...
To find the total inductance of the combination, we need to consider the individual inductances and the mutual inductance between the two coils.
Given data:
Self-inductance of coil 1 (L1) = 100 mH
Self-inductance of coil 2 (L2) = 200 mH
Mutual inductance between the coils (M) = 50 mH
The total inductance of the combination can be calculated using the formula for inductance in parallel aiding order:
1/L = 1/L1 + 1/L2 + 2/M
Step 1: Convert the given values to the same unit (Henry).
L1 = 100 mH = 0.1 H
L2 = 200 mH = 0.2 H
M = 50 mH = 0.05 H
Step 2: Substitute the values into the formula.
1/L = 1/0.1 + 1/0.2 + 2/0.05
Step 3: Simplify the equation.
1/L = 10 + 5 + 40
1/L = 55
Step 4: Take the reciprocal of both sides of the equation.
L = 1/55
Step 5: Convert the result back to millihenries.
L = 1/55 H = 18.18 mH
Therefore, the total inductance of the combination is approximately 18.18 mH.
None of the given options match the calculated value, which suggests that there may be an error in the question or the provided answer options.
Given data:
Self-inductance of coil 1 (L1) = 100 mH
Self-inductance of coil 2 (L2) = 200 mH
Mutual inductance between the coils (M) = 50 mH
The total inductance of the combination can be calculated using the formula for inductance in parallel aiding order:
1/L = 1/L1 + 1/L2 + 2/M
Step 1: Convert the given values to the same unit (Henry).
L1 = 100 mH = 0.1 H
L2 = 200 mH = 0.2 H
M = 50 mH = 0.05 H
Step 2: Substitute the values into the formula.
1/L = 1/0.1 + 1/0.2 + 2/0.05
Step 3: Simplify the equation.
1/L = 10 + 5 + 40
1/L = 55
Step 4: Take the reciprocal of both sides of the equation.
L = 1/55
Step 5: Convert the result back to millihenries.
L = 1/55 H = 18.18 mH
Therefore, the total inductance of the combination is approximately 18.18 mH.
None of the given options match the calculated value, which suggests that there may be an error in the question or the provided answer options.
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Community Answer
Two coils with self-inductance of 100 mH and 200 mH and mutual inducta...
Concept:
The equivalent inductance of series aiding connection is
L = L1 + L2 + 2MThe equivalent inductance of series opposing connection is
L = L1 + L2 – 2M
Equivalent inductance of parallel aiding connection is
Equivalent inductance of parallel opposing connection is
Calculation:
Given,
L1 = 100 mH
L2 = 200 mH
M = 50 mH
Inductance of parallel aiding connection is
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Two coils with self-inductance of 100 mH and 200 mH and mutual inductance of 50 mH are connected in parallel aiding order. What will be the total inductance of the combination?a)250 mHb)4c)75 mH3.175 mHd)87.5 mHCorrect answer is option 'D'. Can you explain this answer?
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