Given the inductor of inductance 5mH, 10mH and 20mH connected in series, the effective inductance is
- 0.35mH
- 3.50mH
- 2.90mH
-
35.0mH
The correct answer is: D
Explanation
Left = L1 + L2 + L3
= 5mH + 10mH + 20mH = 35mH
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