The general form of rectangular notation for representing complex impedances is:
\[Z=R \pm jX\]
Where R is the resistance component, and X (indicated by the j) is the reactance component.
A negative quantity for the reactance term represents a capacitive reactance. (memory aid: the minus sign is straight like the 2 parallel lines in a capacitor schematic)
A positive quantity for the reactance term represents an inductive reactance. (memory aid: an inductor is a coil of wire and not straight like the minus sign that indicates capacitive reactance)
Therefore the rectangular form of a 100Ω purely capacitive reactance is will have:
Zero Resistance: \(R = 0\)
100Ω Capacitive Reactance: \(X = -j100\)
In Rectangular form: \(0 - j100 Ω\)
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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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The "Y-axis" is the vertical axis on a plot. The word "scale" is a big hint, and it rules out "Scatter". A "scatter plot" places dots where an X value produces a Y value. it's a drawing style of plot, not a scale. Scale really refers to the numerical values along the Y axis.
The idea of a "random" scale is just funny. The point of any XY plot is to visualize relationships between the X and Y values. A random scale would make that much harder.
So, lets think about linear vs logarithmic. For linear, you can think of percent or just decimal numbers. For logarithmic, you can think of dB. It's very common to have filters that attenuate by -10, -20, -30, -40dB or more. On a logarithmic plot, those are all separated by the same distance. But on a linear plot, those are 0.1, 0.01, 0.001, and 0.0001. If the linear plot goes from 0 to 1, it's going to be very hard to distinguish those last few. Whenever signals vary over huge ranges (the biggest is many times the smallest), you're going to want a logarithmic scale.
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Phase relationship is the key in this question to remember the answer is phasor diagram.
Think... Phase=Phasor
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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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Electrical impedance can be represented as a complex number of the form \(R + jX\), where \(R\) is resistance, \(X\) is reactance, and \(j\) is the imaginary unit (\(\sqrt{-1}\)). This can be graphed on 2 axes, one with the real part (resistance), and one with the imaginary part (reactance). By convention, the real (resistance) axis is horizontal and the imaginary (reactance) axis is vertical. Thus, a pure resistance (\(R + j0\)) would be plotted on the horizontal axis.
Hint: "Horizontal" and "resistance" have the same amount of letters.
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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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Electrical impedance can be represented as a complex number of the form \(R + jX\), where \(R\) is resistance, \(X\) is reactance, and \(j\) is the imaginary unit (\(\sqrt{-1}\)). This can be graphed on 2 axes, one with the real part (resistance), and one with the imaginary part (reactance). By convention, the real (resistance) axis is horizontal and the imaginary (reactance) axis is vertical.
Remember: When impeded by evil, the X Men resist.
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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.
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.
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.
Mnemonic: One rider is over two pedals (pi), two feet frequency), and two legs (L) or calves (C).
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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.
Memorization Tip: 3.505 mhz starts with the number 3, therefore, "Point 3" is the answer.
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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 (106) 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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