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I am trying to find the total impedance of this circuit but im not getting the right answer. According to my notes the total impedance is 1002 Ω, -86.6°.

This is the method i tried: I got

Xc = 111.1, -90° => 0-j111.1

Xl= 125, 90° => 0+j125

R = 60+j0

then from here ive tried multiple different methods. ive done ((Xc*Xl)/(Xc+Xl))+R. Just adding Xc+Xl+R then converting them back to polar form. Nothing i try give me the right answer. What am i doing wrong here?

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  • \$\begingroup\$ So I get 1002 at -86.6 deg for ((Xc*Xl)/(Xc+Xl))+R. Is this not what you got, or is this what you have and it doesn't match the answer key? \$\endgroup\$
    – MikeP
    Commented Aug 4, 2016 at 14:41
  • \$\begingroup\$ 1002 is what it says in the answer key. Thats not what im getting. \$\endgroup\$
    – Joo223
    Commented Aug 4, 2016 at 14:44

1 Answer 1

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According to my notes the total impedance is 1002 Ω, -86.6°

Your notes are correct and I think the mistake you may have made is adding 60 ohm (resistive) to the 1000 ohms (reactive) to get 1060 ohms. You need to add them as squares then take the square root hence \$\sqrt{1000^2 + 60^2}\$ = 1001.8 ohms.

The 1000 ohms is the reactive impedance of C in parallel with L. I'm presuming you know how to derive this because you appear to be making the right noises in your question i.e. the "(Xc*Xl)/(Xc+Xl)" bit is correct.

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  • \$\begingroup\$ But when i do, (Xc*Xl)/(Xc+Xl) aka (111.1*125)/(111.1+125) im getting 58.82 as answer. \$\endgroup\$
    – Joo223
    Commented Aug 4, 2016 at 14:46
  • \$\begingroup\$ The denominator is a subtraction not an addition i.e. 125 - 111.1111. This is because of the -j for the capacitor. \$\endgroup\$
    – Andy aka
    Commented Aug 4, 2016 at 14:48

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