Question

In the given circuit below inductance values of \(L_1, L_2\) and \(L_3\) are same. The magnetic energy stored in the entire circuit is \((U_t)\) and that stored in the \(L_2\) inductor is \((U_j)\). \(U_t / U_j\) is ____. (Ignore the mutual inductance if any).

Answer 1

Correct Answer: 6

Step 1: Understanding the Question:
The circuit consists of \(L_1\) in series with a parallel combination of \(L_2\) and \(L_3\). We need to find the ratio of total energy to the energy in \(L_2\). All inductances are equal (\(L_1 = L_2 = L_3 = L\)).
Step 2: Key Formula or Approach:
1. Energy stored in an inductor: \(U = \frac{1}{2}LI^2\).
2. Equivalent inductance of parallel inductors: \(L_p = \frac{L \times L}{L + L} = \frac{L}{2}\).
3. Equivalent inductance of series combination: \(L_{eq} = L_1 + L_p\).
Step 3: Detailed Explanation:
Let the total current entering the circuit be \(I\). This current \(I\) passes through \(L_1\).
At the junction, the current splits between \(L_2\) and \(L_3\). Since their inductances are equal, the current divides equally:
\[ I_{L2} = I_{L3} = \frac{I}{2} \]
Total equivalent inductance \(L_{eq}\):
\[ L_{eq} = L + \frac{L \times L}{L + L} = L + \frac{L}{2} = \frac{3L}{2} \]
Total energy stored in the circuit (\(U_t\)):
\[ U_t = \frac{1}{2} L_{eq} I^2 = \frac{1}{2} \left(\frac{3L}{2}\right) I^2 = \frac{3}{4} LI^2 \]
Energy stored in inductor \(L_2\) (\(U_j\)):
\[ U_j = \frac{1}{2} L (I_{L2})^2 = \frac{1}{2} L \left(\frac{I}{2}\right)^2 = \frac{1}{2} L \cdot \frac{I^2}{4} = \frac{1}{8} LI^2 \]
Ratio \(U_t / U_j\):
\[ \frac{U_t}{U_j} = \frac{\frac{3}{4} LI^2}{\frac{1}{8} LI^2} = \frac{3}{4} \times 8 = 6 \]
Step 4: Final Answer:
The ratio \(U_t / U_j\) is \(6\).