Current Growth in an LR Circuit

An inductor hates sudden change. Any change in current through it changes the magnetic flux it stores, and that changing flux induces a back emf that opposes the change, by Faraday's and Lenz's laws. The instant a switch closes on a battery, resistor, and inductor in series, the circuit wants current to jump straight to V/R. But jumping instantaneously means an infinite dI/dt, which would require an infinite back emf. Nothing supplies that, so the current instead rises gradually, easing toward its final value.

Apply the loop rule at any instant: V - IR - L(dI/dt) = 0. This is a first-order differential equation in I(t). Solving it with the initial condition I(0) = 0 gives I(t) = (V/R)(1 - e^−tR/L). At t = 0 the exponential is 1, so I = 0, matching the inductor's refusal to allow instant current. As t grows large, the exponential dies off and I approaches V/R, the steady current set purely by the resistor once the inductor stops fighting. The quantity L/R is the time constant: it tells you how fast the climb happens, the same role RC played for capacitors charging.

While current is still building, the voltage across the inductor is V_L = L(dI/dt) = Ve^−tR/L. It starts at the full battery voltage V, when the inductor is fighting hardest, and decays to zero as the current levels off and the resistor takes over the full voltage drop. This growth behavior sets up its mirror image: once current is flowing and the switch is used to remove the battery, the same differential equation structure governs current decay, just with the exponential now pulling I down toward zero instead of up toward V/R.