LR circuit

Also called RL circuit, Resistor-inductor circuit

An LR circuit contains a resistor and an inductor, so its current needs time to settle. The instant a switch is thrown the inductor blocks the current entirely, and after a long time it behaves as a plain wire, which is the reverse of what a capacitor does.

Topic 13.5 of AP Physics C: Electricity and Magnetism, and of that course only. Apply Kirchhoff's loop rule to a series loop of battery, resistor and inductor and you get the derived equation the CED prints at 13.5.A.2:

E=IR+LdIdt\mathcal{E} = IR + L\frac{dI}{dt}

That equation is not on the equation sheet. The time constant τ=L/Req\tau = L / R_{\text{eq}} is.

The two limits are what questions turn on.

  • At the switching instant. The induced emf is equal in magnitude and opposite in direction to the applied potential difference across the branch containing the inductor (13.5.A.4.i). Nothing is left over to drive current, so the inductor branch carries none: treat it as a break.
  • After a time much greater than τ\tau. An inductor behaves as a conducting wire with zero resistance (13.5.A.4.iii). The current has stopped changing, so there is nothing to oppose.

Those are the mirror image of an RC circuit, where the capacitor starts as a wire and ends as a break. Getting the two the wrong way round is the standard error.

The CED also fixes what one τ\tau buys: for an inductor with zero initial current it is the time to reach roughly 63 percent of the final current, and for one with an initial current it is the time to fall to roughly 37 percent of it. See time constant for both circuits side by side.

As the current changes, the resistor dissipates energy that was stored in the inductor (13.5.A.1), which is where UL=12LI2U_L = \tfrac{1}{2}LI^2 goes.

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