What is the impedance of a network composed of a
100-picofarad capacitor in parallel with a
4000-ohm resistor, at
500 KHz?

Specify your answer in polar coordinates.

2490 ohms, /-51.5 degrees

For more information, please see Ham Radio School site article called Complex Impedance Part 3: Putting It All Together.

Please see Web Archive Org site for the article Parallel RC Circuits The Circuit

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In polar coordinates, what is the impedance of a network composed of a
100-ohm-reactance inductor in series with a
100-ohm resistor?

141 ohms, /45 degrees

From wp2ahg:

\begin{align} Z &= \sqrt{(R^2 + (XL- XC)^2)}\\ &= \sqrt{(100^2\text{ ohms} + (100\text{ ohms} - 0\text{ ohms})^2)}\\ &=\sqrt{(10,000\text{ ohms} + 10,000\text{ ohms}}\\ &=\sqrt{(20,000\text{ ohms}}\\ &= 141\text{ ohms}\\ \\ \end{align}

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In polar coordinates, what is the impedance of a network composed of a
400-ohm-reactance capacitor in series with a
300-ohm resistor?

500 ohms, /-53.1 degrees

From wp2ahg:

(Xl - XR) tells you if it's positive or negative.
(0 ohms - 300 ohms) = -300 ohms, so it's negative.

\begin{align} Z &= \sqrt{(R^2 + (XL- XC)^2)}\\ &= \sqrt{(300^2\text{ ohms} + (400\text{ ohms} - 0\text{ ohms})^2)}\\ &=\sqrt{(900\text{ ohms} + 1,600\text{ ohms}}\\ &=\sqrt{(2,500\text{ ohms}}\\ &= 500\text{ ohms}\\ \\ \end{align}

So, the answer is 500 ohms, and negative.

Answer C is the only 500 ohm/negative answer, so that's the right choice.

You can calculate the degrees if you want, but it's not necessary for answering this question.

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In polar coordinates, what is the impedance of a network composed of a
300-ohm-reactance capacitor, a
600-ohm-reactance inductor, and a
400-ohm resistor,
all connected in series?

500 ohms, /37 degrees

From wp2ahg:

\begin{align} Z &= \sqrt{(R^2 + (XL- XC)^2)}\\ &= \sqrt{(400^2\text{ ohms} + (600\text{ ohms} - 300\text{ ohms})^2)}\\ &=\sqrt{(160,000\text{ ohms} + 90,000\text{ ohms}}\\ &=\sqrt{(250,000\text{ ohms}}\\ &= 500\text{ ohms}\\ \\ \end{align}

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In polar coordinates, what is the impedance of a network comprised of a
400-ohm-reactance inductor in parallel with a
300-ohm resistor?

240 ohms, /36.9 degrees

From wp2ahg:

Total impedance for a parallel RL circuit is:
    \[{\text{Impedance}=\frac{\text{Resistance x Reactance}}{\sqrt{\text{Resistance^2} \ + \text{Reactance^2}}\\}}\]

\[{\text{Impedance}=\frac{300 * 400}{\sqrt{300^2 \ + 400^2}\\}}\]

\[{\text{Impedance}=\frac{300 * 400}{\sqrt{90,000 \ + 160,000}\\}}\]

\[{\text{Impedance}=\frac{120,000}{\sqrt{250,000}\\}}\]

\[{\text{Impedance}=\frac{120,000}{{500}\\}}\]

\[{\text{Impedance}={240\text{ ohms}\\}}\]

Phase Angle for a parallel RL circuit is
   = Degrees(arctan(Reactance / Resistance))°

   = Degrees(arctan(300/400))°
   = Degrees(arctan(0.75))°
   = Degrees(0.64)°
   = 36.9° degrees ◔

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Using the polar coordinate system, what visual representation would you get of a voltage in a sine wave circuit?

The plot shows the magnitude and phase angle.

For well-illustrated explanation, please see Wikipedia's article Bode plot

Also, please see Resources PCB Cadence site for the article Interpreting the Phase in a Bode Plot

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