Question

In the below diagram inductance \(L_1 = L_2 = L_3 = L\). Find the ratio of total energy stored in all inductors to energy stored in inductor \(L_2\). 

Answer 1

Correct Answer: 6

Concept: Energy stored in an inductor: \[ E=\frac{1}{2}LI^2 \] 

Step 1: Circuit analysis Given \[ L_1 = L_2 = L_3 = L \] Current \(I\) flows through \(L_1\). After that the current splits equally into the two identical inductors \(L_2\) and \(L_3\). \[ I_{L_2}=I_{L_3}=\frac{I}{2} \] Step 2: Equivalent inductance Parallel combination of \(L_2\) and \(L_3\): \[ L_p=\frac{L \times L}{L+L}=\frac{L}{2} \] Thus total equivalent inductance \[ L_{eq}=L_1 + L_p \] \[ L_{eq}=L+\frac{L}{2}=\frac{3L}{2} \] Step 3: Total stored energy \[ E_{total}=\frac12 L_{eq} I^2 \] \[ E_{total}=\frac12 \left(\frac{3L}{2}\right)I^2 \] \[ E_{total}=\frac{3}{4}LI^2 \] Step 4: Energy in \(L_2\) Current through \(L_2 = \frac{I}{2}\) \[ E_{L_2}=\frac12 L \left(\frac{I}{2}\right)^2 \] \[ E_{L_2}=\frac{1}{8}LI^2 \] Step 5: Required ratio \[ \frac{E_{total}}{E_{L_2}} = \frac{\frac{3}{4}LI^2}{\frac{1}{8}LI^2} \] \[ =6 \]