Question

When three inductors of same inductance '\(L\)' are connected in series and '\(I\)' is the current passing through the circuit. The energy stored in the circuit is

• A. \(\frac{1}{2}\text{LI}^2\)
• B. \(\frac{3}{2}\text{LI}^2\)
• C. \(\frac{5}{2}\text{LI}^2\)
• D. \(\frac{7}{2}\text{LI}^2\)

Hint:

Series inductors add directly like resistors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When inductors are connected in series, their equivalent inductance adds up, just like resistors in series (assuming no mutual inductance).
The total magnetic energy stored in an inductive circuit depends on the equivalent inductance and the current flowing through it.
Step 2: Key Formulas or Approach:
Equivalent inductance in series: \( L_{\text{eq}} = L_1 + L_2 + L_3 + \dots \)
Energy stored in an inductor: \( U = \frac{1}{2} L_{\text{eq}} I^2 \).
Step 3: Detailed Explanation:
We are given three inductors, each with an inductance of \( L \).
They are connected in series.
The equivalent inductance of the combination is:
\[ L_{\text{eq}} = L + L + L = 3L \] A steady current \( I \) passes through the entire series combination.
The total energy stored in the circuit is calculated using the equivalent inductance:
\[ U = \frac{1}{2} L_{\text{eq}} I^2 \] Substitute \( L_{\text{eq}} = 3L \) into the formula:
\[ U = \frac{1}{2} (3L) I^2 \] \[ U = \frac{3}{2} L I^2 \] Alternatively, energy is scalar and additive. Each inductor stores \( \frac{1}{2}LI^2 \). Total energy = \( 3 \times (\frac{1}{2}LI^2) = \frac{3}{2}LI^2 \).
Step 4: Final Answer:
The energy stored in the circuit is \( \frac{3}{2}\text{LI}^2 \).