In a parallel circuit, the resistive part will always be smaller than the resistance, so less than 4000 ohms. The reactance will be negative if it's capacitive. Only one answer is less than 4000 ohms and has a negative reactance, the one with the capacitor.

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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The resistive and reactive components are both 100 ohms, so you'll get a 45 degree angle - just look for that. You don't have to compute anything.

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, the magnitude is always going to be larger with the reactance added to the resistance, so look for something greater than the resistance. That leaves two choices. The angle is always negative for capacitors, so look for the minus sign. No computation is required.

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, the magnitude is always going to be larger than the resistance - that leaves two choices. The reactance is going to be either + for inductance or - for capacitance. In this case, the inductive reactance is 300 ohms greatr than the capacitive reactance, so net inductive, 300 ohms inductive. Inductive reactance is always positive. No calculations are required.

The 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, the magnitude of a parallel circuit is always going to be less than the resistor. The angle is going to be positive if it's inductive. There is only one choice that's less than 300 ohms and positive. No calculations are needed.

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(400/300))°
   = Degrees(arctan(1.33))°
   = Degrees()°
   = 53.1° degrees ◔

   = 90° (Inductance) - 53.1° = 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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