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 _______.

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: \(4/3\)