An inductor of inductance 10 mH having resistance of 100 , 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 _______.

Step-by-step solution

In an LR circuit, the equation for the voltage is given by Kirchhoff's loop law: E = L di dt + iR The instantaneous voltage across the ideal inductor is the induced EMF, which is given by: V_L = L di dt = E - iR Given values are: E = 1.0 V R = 100 , For current i₁ = 2 mA = 2 10⁻³ A, the voltage across the inductor is: V_ L1 = 1.0 - (2 10⁻³ 100) = 1.0 - 0.2 = 0.8 V For current i₂ = 4 mA = 4 10⁻³ A, the voltage across the inductor is: V_ L2 = 1.0 - (4 10⁻³ 100) = 1.0 - 0.4 = 0.6 V The ratio of the instantaneous volta