An inductor of inductance 10 mH having resistance of 100 $$\Omega$$ is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is _______.

Given parameters for the series RL circuit:

Inductance, $$ L = 10 \text{ mH} $$

Resistance, $$ R = 100 \ \Omega $$

Battery E.M.F., $$ E = 1.0 \text{ V} $$

According to Kirchhoff's Voltage Law (KVL)

$$ E = V_R + V_L $$

Where:

$$ V_R $$ is the voltage across the resistor ($$ V_R = iR $$)

$$ V_L $$ is the instantaneous voltage across the inductor

Rearranging the equation to solve for $$ V_L $$:

$$ V_L = E - iR $$

$$ V_{L1} = E - i_1 R $$

$$ V_{L1} = 1.0 - (2 \times 10^{-3})(100) $$

$$ V_{L1} = 1.0 - 0.2 $$

$$ V_{L1} = 0.8 \text{ V} $$

$$ V_{L2} = E - i_2 R $$

$$ V_{L2} = 1.0 - (4 \times 10^{-3})(100) $$

$$ V_{L2} = 1.0 - 0.4 $$

$$ V_{L2} = 0.6 \text{ V} $$

Now, we calculate the ratio of these instantaneous voltages:

$$ \text{Ratio} = \frac{V_{L1}}{V_{L2}} $$

$$ \text{Ratio} = \frac{0.8 \text{ V}}{0.6 \text{ V}} $$

$$ \text{Ratio} = \frac{8}{6} $$

$$ \text{Ratio} = \frac{4}{3} $$