Linear Circuits 1 - 27 - Inductors, Part 3

Analyze the first circuit, more complex than prior resistor-inductor circuits, using the process to determine the voltage drop at t = 0 s with a voltage source.

Find i(t=0−) for the inductor as the switch opens. Treat the inductor as a short at 0−; apply current division to yield 240 μA through the inductor.

Shows that the inductor current cannot change instantly; the initial current i0 equals the pre-switch value and remains the same after the switch opens.

Observe the inductor acting as a short circuit over time and the current decaying to zero after the switch opens, starting at 240 microamps.

Just after the switch opens, V_A - V_B = -720 mV, so the inductor acts as a 720 mV source; later, the current decays to zero and voltage is zero.

Zero the sources, determine the resistance the inductor current sees, with L = 30 mH and R = 3 kΩ, yielding a time constant of 10 microseconds.

Figure the inductor voltage form after the switch opens using the initial zero volts and a -720 mV source with a 10 μs time constant, yielding a decaying exponential.

Compute the inductor current i(t=0−) just before the switch opens and apply Ohm's law, equivalent resistance, and current division to analyze post-switch currents.

Identify the initial inductor current at t=0+, using the rule that inductor current cannot change instantly and equals its value just before the switch opens.

As time goes to infinity, the inductor becomes a short circuit. The current decays to zero because the circuit is disconnected from the voltage source and no energy sources remain.

Compute the inductor voltage at t=0+ and as t→∞; the current cannot change instantly, so V_L jumps to -1.01 V at t=0+ and then decays to zero.

Zero out voltage sources and compute the series resistance to find tau, the inductor's time constant; with L=6 nH and R=8 ohms, current cannot change instantly, tau=750 ps.

Choose an equation and plug in values to model the inductor voltage after switch opens, showing a jump to -1.01 volts and current decay with a 750 picoseconds time constant.

Apply the same step-by-step process to problems, trust the process, and anticipate more examples in the next lesson.