Resister inductor - calculate resulting impedance

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Discussion Overview

The discussion centers around calculating the resulting impedance of a 600Ω resistor in parallel with a 300jΩ inductor. Participants are exploring the mathematical process behind this calculation, including the representation of complex impedances.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant asks how the expression 600Ω || 300jΩ results in 120 + j240, seeking clarification on the calculation method.
  • Another participant inquires if the original poster is asking about the equivalent resistance of the given components in parallel.
  • There is a suggestion for the original poster to check the equation for combining impedances in parallel and to share their findings.
  • Some participants express frustration over the lack of effort shown by the original poster in attempting to solve the problem independently.
  • Clarification is requested regarding the terminology used, specifically correcting "register" to "resistor." Questions about the original poster's familiarity with electricity and complex numbers are raised.
  • A link to a Wikipedia article on parallel resistor combinations is provided as a resource for further understanding.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the calculation method, and there are multiple competing views regarding how to approach the problem. The discussion remains unresolved.

Contextual Notes

There are indications of missing assumptions regarding the original poster's understanding of complex numbers and impedance calculations, as well as a lack of clarity on the specific steps needed to arrive at the result of 120 + j240.

xdeimos
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resister // inductor -- calculate resulting impedance

Hello,

how 600Ω || 300jΩ = 120 + j240 ??

how do you get the value 120 + j240 from 600 register parallel wih 300jΩ inductor?
 
Last edited:
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Are you tying to ask how to calculate the equivalent resistance of a 600-Ohm resistor in parallel with a 300j Ohm inductor?
 
Last edited by a moderator:
xdeimos said:
Hello,

how 600Ω || 300jΩ = 120 + j240 ??

how do you calculate this

Welcome to the PF.

(I've corrected your text speak "u" in several of your threads -- please check your PMs.)

Are you familiar with the equation for combining impedances in parallel? Maybe try checking wikipedia for this equation, and post what you find here. :smile:
 
stream king , yes that is my question
 
xdeimos said:
stream king , yes that is my question

And? Please show some effort on your part, or this thread will be deleted. We do not spoon feed students here with information that they should be learning on their own. We are here to help after you have made an effort on your own, show those efforts here, and are stuck or confused.
 
?
600Ω || 300jΩ = 120 + j240 ??
I want to know how 600Ω || 300jΩ is equal with 120 + j240 ?
what do i have to explain about?
how 600 register parallel with 300johm is equal to 120+ j240??
 
Well, for starters, stop calling it 'register'. It's a 'resistor'. Do you know anything about electricity?
 
xdeimos said:
?
600Ω || 300jΩ = 120 + j240 ??
I want to know how 600Ω || 300jΩ is equal with 120 + j240 ?
what do i have to explain about?
how 600 register parallel with 300johm is equal to 120+ j240??

Parallel resistor combinations: http://en.wikipedia.org/wiki/Resistor#Series_and_parallel_resistors

Now, are you familiar with complex numbers and how impedances are represented?
 

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