Question

When four inductors of the same inductances L are connected in series, and I is the current passing through the circuit. The energy stored in the circuit is

• A. \(LI^{2}\)
• B. \(2LI^{2}\)
• C. \(3LI^{2}\)
• D. \(4LI^{2}\)

Hint:

In series the inductances add, and energy is half times L times current squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Inductors in series add: \(L_{eq}=L+L+L+L=4L\).

Step 2: Energy
\(U=\dfrac{1}{2}L_{eq}I^2=\dfrac{1}{2}(4L)I^2\).

Step 3: Result
\(U=2LI^2\).

Final Answer:
The energy stored is 2LI squared. \[ \boxed{\text{(B) }2LI^{2}} \]