Question
Download Solution PDFThe self-inductances of three coils are LA = 20H, LB = 30H and LC = 40 H. The coils are connected in series in such a way that fluxes of LA and LB add, fluxes of LA and LC are in opposition and fluxes of LB and LC are in opposition. If MAB = 8H, MBC = 12H and MAC = 10H, what is the total inductance of the circuit?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The equivalent inductance of series-aiding connection is
Leq = L1 + L2 + 2M
The equivalent inductance of a series opposing connections is
Leq = L1 + L2 – 2M
The equivalent inductance of parallel aiding connection is
Leq = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} - 2M}}\)
Equivalent inductance of parallel opposing connection is
Leq = \(\frac{{{L_1}{L_2} - {M^2}}}{{{L_1} + {L_2} + 2M}}\)
Calculation:
Total inductance (LT) = LA + LB + LC + 2 MAB - 2 MBC - 2 MAC
LT = 20 + 30 +40 + 2(8) - 2(12) - 2(10) = 62 H
Last updated on Jul 2, 2025
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