An inductor of inductance 10 mH having resistance of 100\, is...

In an LR circuit, the equation for the voltage is given by Kirchhoff's loop law:

$E = L \dfrac{di}{dt} + iR$

The instantaneous voltage across the ideal inductor is the induced EMF, which is given by:

$V_L = L \dfrac{di}{dt} = E - iR$

Given values are:
$E = 1.0$ V
$R = 100\,\Omega$

For current $i_1 = 2$ mA $= 2 \times 10^{-3}$ A, the voltage across the inductor is:

$V_{L1} = 1.0 - (2 \times 10^{-3} \times 100) = 1.0 - 0.2 = 0.8$ V

For current $i_2 = 4$ mA $= 4 \times 10^{-3}$ A, the voltage across the inductor is:

$V_{L2} = 1.0 - (4 \times 10^{-3} \times 100) = 1.0 - 0.4 = 0.6$ V

The ratio of the instantaneous voltages is:

$\dfrac{V_{L1}}{V_{L2}} = \dfrac{0.8}{0.6} = \dfrac{4}{3}$

Answer: $4/3$