A $100 - \Omega$ resistor is connected in series with a $10.0 - \mathrm { mH }$ inductor across an ac source operating at $1.00 \mathrm { kHz }$. What is the impedance of this circuit?
A) $1000 \Omega$
B) $236 \Omega$
C) $200 \Omega$
D) $118 \Omega$
E) $100 \Omega$

Question

Quizplus · Accepted Answer

The impedance of an inductor is given by $XL = 2\pi fL$, where $f$ is the frequency and $L$ is the inductance. The impedance of a resistor is given by $R$. The impedance of a series circuit is given by $Z = \sqrt{R^2 + XL^2}$. Plugging in the given values, we get $Z = \sqrt{100^2 + (2\pi \times 1000 \times 0.01)^2}$ = 118 $\Omega$.

A 100−Ω100 - \Omega100−Ω Resistor Is Connected in Series with A 10.0−mH10.0 - \mathrm { mH }10.0−mH

A $100 - \Omega$ Resistor Is Connected in Series with A $10.0 - \mathrm { mH }$