A 1,000 $$Omega$$ resistor is in series with a 2-mH inductor.  An ac voltage source operating at a frequency of 100,000 rad/s is attached as shown in the figure. The impedance ($$Omega$$) of the RL combination is most nearly:

A.   200 + j1,000
B.   1,000 + j200
C.   38.4 + j192
D.   1,000 - j200

The question asks us to find the total impedance of the RL combination. As these components are in series with each other, we can find the total impedance by adding together their individual impedances, as shown below:

$Z_{RL}=Z_R+Z_L$

The impedance of the resistor is real and is equal to its total resistance. Therefore:

$Z_R=1000\Omega$

The impedance of an inductor is purely imaginary and depends on its inductance as well as frequency, according to the following formula:

$Z_L=j\omega L$

The problem tells us that the inductance (L) of the inductor is 2mH, and the angular frequency ($\omega$) across it is equal to 100,000 rad/s. Therefore:

$$\begin{gathered} Z_L=j(100,000rad/s)(.002H) \\ {Z}_L=j200\Omega \end{gathered}$$

Now that we have the individual impedances, we can write the total impedance:

$Z_{RL}=(1000+j200)\Omega$

Our answer is B.