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  4. A resistance of 40 Ω and an inductance of 95.5 mH are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly:

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Q. A resistance of $40 \Omega$ and an inductance of $95.5 mH$ are connected in series in a 50 cycle/second ac circuit. The impedance of this combination is very nearly:

Alternating Current

Solution:

$Z=\sqrt{R^{2}+X_{L}^{2}}=\sqrt{R^{2}+(2 \pi v L)^{2}}$
$Z=\sqrt{(40)^{2}+4 \pi^{2} \times(50)^{2} \times\left(95.5 \times 10^{-3}\right)^{2}}$
$=50 \Omega$