Two different coils have self inductances `L_(1)=8mH` and `L_(2)=2mH`. The current in both the coil is increased at same constant rate. At a certain instant power given to two coils is same. At that time the energy stored in both the coils are `V_(1) & V_(2)` respectively, then `(V_(1))/(V_(2))` is

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. 2
D. 4

Answer

To solve the problem, we need to find the ratio of the energy stored in two coils with different self-inductances when the power supplied to both coils is the same and the current is increased at the same constant rate.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Self-inductance of coil 1, \( L_1 = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H} \)
- Self-inductance of coil 2, \( L_2 = 2 \, \text{mH} = 2 \times 10^{-3} \, \text{H} \)

2. **Understand the Power Relation**:
- The power \( P \) delivered to an inductor is given by:
\[ P = E \cdot I \]
where \( E \) is the induced EMF and \( I \) is the current.

3. **Induced EMF Formula**:
- The induced EMF for each coil can be expressed as:
\[ E_1 = -L_1 \frac{dI_1}{dt} \quad \text{and} \quad E_2 = -L_2 \frac{dI_2}{dt} \]

4. **Set Up Power Equations**:
- Since the power is the same for both coils, we can write:
\[ E_1 I_1 = E_2 I_2 \]
- Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ (-L_1 \frac{dI_1}{dt}) I_1 = (-L_2 \frac{dI_2}{dt}) I_2 \]

5. **Since \( \frac{dI_1}{dt} = \frac{dI_2}{dt} \)**:
- We can cancel \( \frac{dI}{dt} \) from both sides:
\[ L_1 I_1 = L_2 I_2 \]

6. **Express the Current Ratio**:
- Rearranging gives:
\[ \frac{I_2}{I_1} = \frac{L_1}{L_2} \]
- Substituting the values of \( L_1 \) and \( L_2 \):
\[ \frac{I_2}{I_1} = \frac{8 \, \text{mH}}{2 \, \text{mH}} = 4 \]

7. **Energy Stored in the Inductors**:
- The energy stored in an inductor is given by:
\[ V = \frac{1}{2} L I^2 \]
- Therefore, for both coils:
\[ V_1 = \frac{1}{2} L_1 I_1^2 \quad \text{and} \quad V_2 = \frac{1}{2} L_2 I_2^2 \]

8. **Find the Ratio of Energies**:
- The ratio \( \frac{V_1}{V_2} \) can be expressed as:
\[ \frac{V_1}{V_2} = \frac{L_1 I_1^2}{L_2 I_2^2} \]
- Substituting \( I_2 = 4 I_1 \):
\[ \frac{V_1}{V_2} = \frac{L_1 I_1^2}{L_2 (4 I_1)^2} = \frac{L_1 I_1^2}{L_2 \cdot 16 I_1^2} \]
- Cancel \( I_1^2 \):
\[ \frac{V_1}{V_2} = \frac{L_1}{16 L_2} \]
- Substituting the values of \( L_1 \) and \( L_2 \):
\[ \frac{V_1}{V_2} = \frac{8 \times 10^{-3}}{16 \cdot 2 \times 10^{-3}} = \frac{8}{32} = \frac{1}{4} \]

### Final Answer:
\[ \frac{V_1}{V_2} = \frac{1}{4} \]