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Subelement B

Electrical Math

Section 17

Impedance Networks-2

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
  • 4000 ohms, /38.5 degrees
  • 5112 ohms, /-38.5 degrees
  • Correct Answer
    2490 ohms, /-51.5 degrees

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?

  • 121 ohms, /35 degrees
  • Correct Answer
    141 ohms, /45 degrees
  • 161 ohms, /55 degrees
  • 181 ohms, /65 degrees

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?

  • 240 ohms, /36.9 degrees
  • 240 ohms, /-36.9 degrees
  • Correct Answer
    500 ohms, /-53.1 degrees
  • 500 ohms, /53.1 degrees

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?

  • Correct Answer
    500 ohms, /37 degrees
  • 400 ohms, /27 degrees
  • 300 ohms, /17 degrees
  • 200 ohms, /10 degrees

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
  • Correct Answer
    240 ohms, /36.9 degrees
  • 500 ohms, /53.1 degrees
  • 500 ohms, /-53.1 degrees

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 sinewave circuit?

  • To show the reactance which is present.
  • To graphically represent the AC and DC component.
  • To display the data on an XY chart.
  • Correct Answer
    The plot shows the magnitude and phase angle.

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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