- #1

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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?

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- Thread starter xdeimos
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- #1

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Hello,

how 600Ω || 300jΩ = 120 + j240 ??

how do you get the value 120 + j240 from 600 register parallel wih 300jΩ inductor?

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- #2

SteamKing

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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?

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- #3

berkeman

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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.

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stream king , yes that is my question

- #5

berkeman

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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.

- #6

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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???????

- #7

SteamKing

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- #8

berkeman

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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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