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 topic is Electromagnetic Induction and energy stored in inductors.
We are asked to find the ratio of the total magnetic energy stored in a combination of three identical inductors to the energy stored in a specific inductor, \(L_2\).
Step 2: Key Formula or Approach:
The energy stored in an inductor is given by the formula \(U = \frac{1}{2}LI^2\).
We will also use the principle of current division for parallel components.
Step 3: Detailed Explanation:
Let \(L_1 = L_2 = L_3 = L\).
Let the total current from the source be \(I\). This entire current \(I\) passes through inductor \(L_1\).
At the junction, the current \(I\) splits to pass through the parallel combination of \(L_2\) and \(L_3\). Since \(L_2 = L_3\), the current divides equally.
Current through \(L_2\) is \(I_{L2} = I/2\).
Current through \(L_3\) is \(I_{L3} = I/2\).
Now, we calculate the energy stored in each inductor:
Energy in \(L_1\): \(U_1 = \frac{1}{2}L(I)^2 = \frac{1}{2}LI^2\).
Energy in \(L_2\): \(U_j = U_2 = \frac{1}{2}L(I_{L2})^2 = \frac{1}{2}L(I/2)^2 = \frac{1}{8}LI^2\).
Energy in \(L_3\): \(U_3 = \frac{1}{2}L(I_{L3})^2 = \frac{1}{2}L(I/2)^2 = \frac{1}{8}LI^2\).
The total energy stored in the circuit (\(U_t\)) is the sum of the energies in all inductors:
\[ U_t = U_1 + U_2 + U_3 = \frac{1}{2}LI^2 + \frac{1}{8}LI^2 + \frac{1}{8}LI^2 \]
\[ U_t = \left(\frac{1}{2} + \frac{1}{4}\right)LI^2 = \frac{3}{4}LI^2 \]
Finally, we find the required ratio \(U_t / U_j\):
\[ \frac{U_t}{U_j} = \frac{\frac{3}{4} LI^2}{\frac{1}{8} LI^2} = \frac{3}{4} \times \frac{8}{1} = 6 \]
Step 4: Final Answer:
The ratio \(U_t / U_j\) is \(6\).