Given the inductor of inductance 5mH, 10mH and 20mH connected in series, the effective inductance is
Attached Figure / Diagram
A
0.35mH
B
3.50mH
C
2.90mH
D
35.0mH
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Left = L1 + L2 + L3
= 5mH + 10mH + 20mH = 35mH
Source: WAEC / JAMB / NECO Classroom Archive (Record #14492)