Two different coils have self inductances `L_1=9mH` and `L_2=2mH`. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coils is the same. At that time, the current the induced voltage and the energy stored in the first coil are `i_1,V_1` and `W_1` respectively.
Corresponding values for the second coil at the same instant are `i_2,V-2` and `W_2` respectively. Then,

Options

A. `i_1/i_2=1/4`
B. `i_1/i_2=4`
C. `W_1/W_2=1/4`
D. `(V_(1))/(V_(2)) =4`

Explanation

To solve this problem, we need to analyze the relationships between the self-inductances, induced voltages, currents, and energies stored in the two coils.

### Given Data:
- Self-inductance of coil 1: \( L_1 = 9 \, \text{mH} = 9 \times 10^{-3} \, \text{H} \)
- Self-inductance of coil 2: \( L_2 = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)
- The current in both coils is increased at the same constant rate.

### Step 1: Induced Voltage
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