What is the conjugate impedance of 30+j40?

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The conjugate impedance of 30 + j40 is 30 - j40. The magnitude of both the original and conjugate impedance is 50 ohms, calculated as the square root of (30² + 40²). The discussion confirms that replacing j with -j gives the conjugate. Understanding conjugate impedance is crucial in AC circuit analysis. This knowledge is essential for effective impedance matching in electrical engineering.
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Urgent! About Impedance

the total ohm of this impedance ( 30+j40 ) is 50ohm.
but wat is the conjugate impedance ?? 30-j40? total ohm is wat?

30+j40 is 50ohm
solution is square root of (302+402)
 
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Welcome to PF!

Hi KnightMars! Welcome to PF! :smile:
KnightMars said:
the total ohm of this impedance ( 30+j40 ) is 50ohm.
but wat is the conjugate impedance ?? 30-j40? total ohm is wat?

30+j40 is 50ohm
solution is square root of (302+402)

Yes, the complex conjugate of 30 + 40 j is 30 - 40j.

(you simply replace j by -j)

The magnitude of the complex conjugate impedance is always the same as the magnitude of the original impedance (in this case, 50 Ω). :wink:
 


ok ths ^^ done my homework ^^ hehe
 
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