Q43. An inductor of inductance 10 mH has internal resistance 100 Ω and is connected to an ideal 1.0 V battery through a switch, as shown in the figure. After the switch is closed, what is the ratio of the instantaneous inductive voltages when the current is 2 mA and 4 mA, respectively?

Question

G1 Physics · Accepted Answer

At any instant after the switch is closed, Kirchhoff's loop rule gives E - iR - V(L) = 0. Therefore, V(L) = E-iR. For i₁ = 2 mA: V(L1) = 1.0 - (2×10⁻³)(100) = 0.8 V. For i₂ = 4 mA: V(L2) = 1.0 - (4×10⁻³)(100) = 0.6 V. Hence, V(L1) ÷ V(L2) = 0.8 ÷ 0.6 = 4 ÷ 3. Correct option: A.