Assume a 120-V, single-phase source is feeding a load of $$7+j12 (Omega)$$ through a line impedance of $$2+j0 (Omega)$$. The magnitude (V) of the voltage drop across the line is most nearly: 

A.  16
B.  27
C.  97
D.  111

From Ohm's law, we know that the total voltage drop across the line will be equal to the current travelling through the line multiplied by its total impedance in Ohm's:

$V = IZ$

To find the total current of the circuit, we first need to find its total impedance. We have two sources of impedance: the load as well as the line. We must add together the impedance of each to find the total impedance of the circuit:

$$\begin{gathered} Z_T=Z_{Load}+Z_{Line} \\ {Z}_T=\left(7+j12\right)+(2+j0) \end{gathered}$$

We know when adding together complex numbers, we simply add the real parts together and separately add together the imaginary parts. Our total impedance is therefore:

$Z_T=9+j12$

Next, we use Ohm's law to solve for current:

$$\begin{gathered} V=I{Z}_T \\ 120+0j = I·(9+j12) \\ \frac{120+0j}{9+j12}=I \end{gathered}$$

Complex numbers are easier to divide when they are written in polar form. Let's convert both the numerator and denominator from rectangular to polar forms:

$$\begin{gathered} Rec\tan gular Form:a+jb \\ Polar Form: c\angle \theta where c = \sqrt{a^2+b^2}, \theta = {\tan }^{-1}\left(\frac{b}{a}\right) \\ For the numerator, 120+0j: \\ c = \sqrt{{120}^2+0^2}=120 \\ \theta = {\tan }^{-1}\left(\frac{0}{120}\right) = 0 \\ 120+0j = 120\angle 0° \\ For the denominator, 9+j12: \\ c = \sqrt{9^2+{12}^2}=\sqrt{225}=15 \\ \theta = {\tan }^{-1}\left(\frac{12}{9}\right) = 53.13 \\ 9+j12 = 15\angle 53.13° \end{gathered}$$

Now that we have our conversions, our new equation for current looks like this:

$\frac{120\angle 0°}{15\angle 53.13°}=I$

When dividing complex numbers in polar form, we divide the constant and subtract the angle:

$\frac{120\angle 0°}{15\angle 53.13°}=8\angle -53.13°=I$

This represents our total current traveling through the circuit. Now, we want to use Ohm's law once more to find the voltage drop specifically across the line. The total current will equal the line current:

$V_{Line}=I_{Line}{Z}_{line}$

Let's first convert our line impedance to polar form:

$$\begin{gathered} 2+0j \\ c = \sqrt{2^2+0^2}=2 \\ \theta = {\tan }^{-1}\left(\frac{0}{2}\right) = 0 \\ 120+0j = 2\angle 0° \end{gathered}$$

Now let's plug in our values:

$V_{Line}=(8\angle -53.13°)( 2\angle 0°)$

When multiplying complex numbers, we multiply the constants and add the angles. Therefore we get:

$V_{Line}=(8\angle -53.13°)( 2\angle 0°) = 16\angle -53.13°$

Since the question asks us for the magnitude, we can ignore the angle. Therefore, our line voltage is equal to 16V, answer A.