Linear Circuits 1 - 28 - Inductors, Part 4

Set up a resistor-inductor circuit driven by a unit step to show t<0 and t>0 behavior, with initial current from the 50 volt source and gradual rise after turn-on.

Calculate i(t=0−) by treating the inductor as a short before the source turns on, and apply Ohm's law with 50 V, 2 Ω, and 6 Ω to get 25 A.

Apply the inductor property that current cannot change instantly to determine i(t=0+): the inductor current remains 25 A immediately after the second voltage source turns on.

Determine the inductor current as t approaches infinity with both sources on, treating the inductor as a short and using Ohm's law to reach 50 amps.

Learn how to find the voltage across the inductor at t = 0 and t = infinity in a circuit, as part of step four in inductors, part 4.

Zero the sources; resistance seen by the inductor is 1.5 Ω from 2 Ω and 6 Ω in parallel, so tau = 3 H / 1.5 Ω = 2 s.

Select the final equation and plug values to compute inductor current, starting at 25 A, ending at 50 A, with a 2 s time constant under a step input.

Apply v = L di/dt to determine the voltage across an inductor from the current’s rate of change. The lecture demonstrates slope-based calculations for voltage drops across inductors.

Explore how energy stored in an inductor follows E = 1/2 L I^2, with current through the inductor determining the stored energy as time goes to infinity.

Combine inductors in series by adding their values, and in parallel by summing reciprocals to find the equivalent inductance, mirroring resistor rules.

Apply the step function to on/off sources, compute inductor voltage from current, and show capacitors invert these ideas, with voltage that cannot change instantly.