Question

By combining three inductances, \(2\) H, \(4\) H and \(6\) H, the minimum value of effective inductance that can be obtained is (in H)

• A. \(12/11\)
• B. \(11/12\)
• C. \(5/12\)
• D. \(12/5\)

Hint:

The smallest effective inductance occurs when all inductors are connected in parallel.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Inductors in series add, and in parallel their reciprocals add, so the parallel combination gives the minimum.

Step 2: Parallel formula
\(\dfrac{1}{L}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}\)

Step 3: Calculation
\(\dfrac{1}{L}=\dfrac{6+3+2}{12}=\dfrac{11}{12}\)
\(L=\dfrac{12}{11}\) H

Final Answer:
The minimum inductance is 12 by 11 henry. \[ \boxed{\text{(A) }\dfrac{12}{11}} \]