Hint:
A negative imaginary (e.g. -jX) number denotes capacitive reactance
A positive imaginary (e.g. +jX) number denotes inductive reactance.
When using rectangular coordinates to graph the impedance of a circuit, the horizontal axis represents the resistive component. When using rectangular coordinates to graph the impedance of a circuit, the vertical axis represents the reactive component.
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Impedances are described by vectors, having an angle (phase angle) and magnitude (amplitude). This is similar to another question in the pool about vectors.
So, the magnitude is the distance from the origin (center of the graph), and the phase angle is the angle (usually from the X-axis). Positive phase angles are impedances with a net inductive reactance. Negative phase angles are reactances with a net capacitive reactance.
Hint: Polar coordinates deal with angles.
Hint: Phased by a Polar Bear.
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Phase angles are measured from 0 to 180 for positive (inductive) and 0 to -180 for negative (capacitive).
Hint: Positively Phased by a Polar Bear.
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Phase angles displayed in polar coordinates are angles above and below the horizontal line. Capacitive reactances are displayed in the 4th quadrant, the one with a positive resistive element (right side of the graph) and a capacitive reactance - they have a negative phase angle, a point below the horizontal line.
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Phase relationship is the key in this question to remember the answer is phasor diagram.
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In any impedance measurement:
HINT: "50-j25" in the question, & "50 ohms resistance in series with 25" which comes in the same order. So it boils down to remembering if it is inductive or capacitive. I sure that you have the capacity to remember capacitive!
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A vector is defined as a quantity that includes magnitude and direction. The direction could also be described as "an angular component." This is another example of a question of definition.
Impedances can be described as a vector in cartesian or polar coordinates, i.e., a magnitude (distance from the origin of the graph) and direction (an angle, compared to the horizontal or X-axis of the graph).
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Polar coordinates are often used to display the phase angle of a circuit containing resistance, inductive and/or capacitive reactance. In a polar-coordinate system, each point on the graph has two values, a magnitude and an angle.
REF: http://www.kb6nu.com/extra-class-question-of-the-day-impedance-plots-and-coordinate-systems/
Hint: (polar)ization is measured with an angle, and phase is measured with an angle.
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On a rectangular coordinate graph, remember that the horizontal axis plots the resistive component and the vertical axis plots the reactive component.
Visual:
When laying horizontal, one is more resistive to get up
When standing vertical, one is more reactive to their surroundings
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When using rectangular coordinates to graph the impedance of a circuit, the vertical axis represents the reactive component and the horizontal axis represents the resistive component.
Fun Hint: Vertical Y axis... getting a rise (up) out of someone is a reaction (Reactance). Horizontal X axis is dragging that someone who resists (Resistance). -KD5JUN
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This goes back to high school days.. Remember when we had to plot lines on chart paper?
Plotting points on the grid involved determining an X value and a Y value. Another term for this is Rectangular Notation. Remember Rectangles = Rectangular with X (horizontal) and Y (vertical) axis.
-KE0IPR
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The horizontal axis represents purely resistive values, so if your answer lies anywhere along this axis, the value is a pure resistance. KD9AGJ
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A Rectangular coordinate system can be used to represent resistance along the X-axis, and the combination of inductive and capacitive reactance on the Y-axis.
If the sum of the reactance components is inductive, it will be in the 1st quadrant, a positive value of Y.
If the sum of the reactive components is capacitive, it will be in the 4th quadrant, below the X-axis, a negative value of Y.
Hint: Coordinate System = graph, graphs are shown in a rectangle or square
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The impedance of a capacitor, denoted by \(X_C\) is: \[X_C= \frac{1}{2\pi fC}\] where:
The impedance of a capacitor has a negative j value. In the example shown we can mostly remove the large scale values by canceling the (\(10^{6}\)) of the MegaHertz and the (\(10^{-12}\)) of the picoFarads. So, we are left with (\(10^{-6}\)) giving an impedance value that is:
\begin{align} \frac{1}{2\pi\cdot14\cdot38\cdot10^{-6}}&\approx 299.2\:Ω\\ &\Rightarrow- j299.2\:Ω \end{align}
In a series circuit with a 400 Ω resistor the total impedance is \(400 - j299.2\:Ω\) which is in the lower right quadrant of the figure at about 400 in the +X direction and about 300 in the -Y direction.
vaughanth
Hint: Since this circuit is dealing with a capacitor (no inductor component), the capacitance magnitude will be negative. Only Point 4 is a negative capacitance. The pure resistance is 400, so that puts the point on 400.
-KE0IPR
Memory tip. If the frequency is 21 MHz or more, use the first number of the capacitor for the answer clue. If the frequency is less than 21 MHz, use the first number of the resistor for the answer clue.
-KM4VOW
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The impedance of the inductor \(Z_L = 2 \pi f L\), where:
The \(j\) value of the impedance of an inductor is positive.
\[Z_L = 2 \pi \times 3.505\text{ MHz} \times 18\ \mu\text{H}\]
Since the frequency here is in MegaHertz and the Inductance in microhenries, then the Mega (\(10^6\)) and micro (\(10^{-6}\)) exponents cancel:
\begin{align} Z_L &= 2 \pi \times 3.505\text{ MHz} \times 18\ \mu\text{H}\\ &= 2 \pi \times 3.505 \times 18\\ &= 396.4\ \Omega \end{align}
In a series circuit with a \(300\ \Omega\) resistor, the total impedance is \(300 + j396.4\ \Omega\) which is in the upper right quadrant of the figure at about \(300\) in the \(+x\) direction and about \(400\) in the \(+y\) direction, corresponding to Point 3 in Figure E5-2.
Test Tip: Because the resistance is \(300\ \Omega\), the only correct possibilities will be found at 300 ohms positive from the origin. Because the reactive component is only inductive, any correct possibility will be found relatively far away from the resistance axis in the positive direction. Point 3 is the only option that fits.
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The impedance of a capacitor, denoted by \(X_C\) is: \[X_C= \frac{1}{2\pi fC}\] where:
The impedance of a capacitor has a negative j value. In the example shown, we can mostly remove the large scale values by canceling the (10^{6}) of the MegaHertz and the (10^{-12}) of the picoFarads. So, we are left with (10^{-6}), giving an impedance value that is: \begin{align} X_C&= \frac{1}{2\pi (21.2\times 10^6 \text{ Hz})(19\times 10^{-12}\text{ F})} \\ &=\frac{1}{2\pi(21.2)(19)\left(10^{-6}\right)}\\ &\approx -j395.1\:\Omega \end{align}
In a series circuit with a 300 Ω resistor, the total impedance is \(300 - j395.1\:Ω\), which is in the lower right quadrant of the figure, at about 300 in the +X direction and about 400 in the -Y direction.
In short, since the problem only specifies a capacitance (and no inductance), only one answer falls on 300 Ω for the resistance (+X) axis and has a negative reactance: Point 1.
Also, seeing that the answer must be in the fourth quadrant, the only choices are Point 1 and Point 4, but only Point 1 is among the answer choices.
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First, we know that the answer is going to be one of the three points that aligns horizontally at 300 ohms (the pure resistive component). Unfortunately, this only eliminates Point 5 as a possible answer, leaving us with Points 1, 3, and 8 as possibilities.
To determine which answer is correct, we need to compute the impedance of both the capacitor and inductor at the frequency given. We'll start with the impedance of the inductor: (Use the \(π\) key on your calculator for better accuracy.) \begin{align} X_L &= 2πfL\\ &= 2π\cdot(24.9\cdot10^6\text{ Hz}) \cdot (0.64\cdot10^{-6}\text{ H}) \\ &\approx 100.13 \:\Omega\ \end{align}
Then the impedance of the capacitor:
\begin{align} X_C &= \frac{1}{2πfC} \\ &= \frac{1}{2π\cdot(24.9 \cdot10^6 \text{ Hz}) \cdot (85\cdot10^{-12}\text{ F})} \\&= \frac{1}{0.0133} \\ &\approx 75.20\:\Omega \end{align}
This tells us that we have an inductive component (positive direction on the impedance axis) of approximately 100 \(\Omega\) and a capacitive component (negative direction on the impedance axis) of 75 \(\Omega\).
The resultant reactive component will be \(25\:\Omega\) of inductive impedance at the frequency given (\(X_L - X_C\approx 100-75=25\:\Omega\)).
This is a point slightly above the resistance axis in the positive direction. Only Point 8 satisfies that condition.
Test Tip: Because the resistance is \(300\ \Omega\), the only correct possibilities will be found at 300 ohms positive from the origin. Because the reactive component is relatively balanced, any correct possibility will be found relatively close to the resistance axis in the positive direction. Point 8 is the only option that fits.
Test "cheat": do the math to understand these Figure E5-2 problems, BUT for the test, the one's digit in the frequency names the point (14 MHz is Point 4), EXCEPT if the question involves BOTH inductance and capacitance you double that one's digit (2 x 4 = 8, so Point 8).
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