In an LCR circuit with $R=3\,\Omega$, $X_L=6\,\Omega$, $X_C=2\,\Omega$, the impedance (in Ω) is:
Answer (integer)
5
Solution
$Z = \sqrt{3^2 + (6-2)^2} = \sqrt{9+16} = 5$ Ω.
About this question
Subject: Physics · Chapter: Alternating Current · Topic: Phasors and Impedance
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