Question

Two different coils have self - inductance \(L_1 = 9\text{ mH}\) and \(L_2 = 3\text{ mH}\). The current in first coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At certain instant of time, the power given to the two coils is same. At that time, there was current and induced voltage in the two coils. At the same instant, the ratio of the energy stored in the first coil to that in second coil is

• A. 1 : 3
• B. 3 : 1
• C. 1 : 9
• D. 9 : 1

Hint:

Energy in inductor: $U = \frac{1}{2}LI^2$

Answer 1

The Correct Option is B

Concept: \[ \text{Power} = VI = L \frac{dI}{dt} \cdot I \]

Step 1: Given same rate of change of current. \[ P \propto L \cdot I \] Since powers are equal: \[ L_1 I_1 = L_2 I_2 \Rightarrow \frac{I_1}{I_2} = \frac{L_2}{L_1} = \frac{3}{9} = \frac{1}{3} \]

Step 2: Energy stored. \[ U = \frac{1}{2} L I^2 \] \[ \frac{U_1}{U_2} = \frac{L_1 I_1^2}{L_2 I_2^2} \] \[ = \frac{9}{3} \times \left(\frac{1}{3}\right)^2 = 3 \times \frac{1}{9} = \frac{1}{3} \]

Step 3: Invert ratio. \[ U_1 : U_2 = 3 : 1 \] Final Answer: Option (B)