Linear Circuits 1 - 27 - Inductors, Part 3

Name: Linear Circuits 1 - 27 - Inductors, Part 3
Rating: 4.9 (25 reviews)

A simple process that can be used to solve any resistor inductor circuit

Created byMark Budnik, Mark M. Budnik

Last updated 6/2020

English

What you'll learn

A simple process to solve any resistor-inductor linear circuit

Course content

4 sections • 17 lectures • 40m total length

Agenda0:03
Review4:16
A Process to Solve Any Resistor-Inductor Circuit1:19
Explore a two-path, step-by-step method to solve any resistor–inductor circuit, including a six-step process for circuits with no sources and an initial-condition approach for source-containing cases.

A First Example1:35
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.
Step 1 - Finding i(t=0-)3:56
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.
Step 2 - Finding i(t=0+)1:06
Step 3 - Finding i(t=infinity)1:48
Observe the inductor acting as a short circuit over time and the current decaying to zero after the switch opens, starting at 240 microamps.
Step 4 - Finding the Voltage Drop Across the Inductor at t =0 and t = infinity6:39
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.
Step 5 - Finding the Time Constant, Tau, of the Circuit1:46
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.
Step 6 - Choose One of the Final Equations, and Plugin Your Values2:02

Step 1 - Finding i(t=0-)3:12
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.
Step 2 - Finding i(t=0+)1:32
Step 3 - Finding i(t=infinity)1:17
Step 4 - Finding the Voltage Drop Across the Inductor at t =0 and t = infinity6:11
Step 5 - Finding the Time Constant, Tau, of the Circuit2:11
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.
Step 6 - Choose One of the Final Equations, and Plugin Your Values1:12

Requirements

High School or College Physics
Calculus 1 Would Be Extremely Helpful

Description

Day 27 of Linear Circuits. Inductors are one of the three passive circuit components (along with resistors and capacitors). However, their operation and behavior is often shrouded in mystery. After seeing how to find the initial and final values of currents and voltages, today, we introduce a solution process that uses this information for any resistor-inductor circuit. That's pretty cool. : )

The material covers all of the lecture material from an twenty-seventh lecture in a traditional, sophomore-level linear circuits class.

Who this course is for:

Beginner Engineering and Physics Students

Linear Circuits 1 - 27 - Inductors, Part 3

What you'll learn

Explore related topics

Course content

Introduction3 lectures • 6min

A First Example7 lectures • 19min

A Second Example6 lectures • 16min

Summary1 lecture • 1min

Requirements

Description

Who this course is for: