Given the inductor of inductance 5mH, 10mH and 20mH connected in series, the effective inductance is
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Correct Answer: Option D
Explanation:
Left = L1 + L2 + L3
= 5mH + 10mH + 20mH = 35mH
Left = L1 + L2 + L3
= 5mH + 10mH + 20mH = 35mH