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Point 2
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Correct AnswerPoint 4
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Point 5
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Point 6
The impedance of a capacitor, denoted by \(X_C\) is: \[X_C= \frac{1}{2\pi fC}\] where:
- \(f\) is the frequency of interest
- \(C\) is the capacitance in Farads.
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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Point 1
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Correct AnswerPoint 3
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Point 7
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Point 8
The impedance of the inductor \(Z_L = 2 \pi f L\), where:
- \(f\) is the frequency of interest
- \(L\) is the inductance in Henries.
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.
Last edited by knguyen553. Register to edit
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Correct AnswerPoint 1
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Point 3
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Point 7
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Point 8
The impedance of a capacitor, denoted by \(X_C\) is: \[X_C= \frac{1}{2\pi fC}\] where:
- \(f\) is the frequency of interest
- \(C\) is the capacitance in Farads.
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.
Last edited by marvsherman419. Register to edit
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Point 1
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Point 3
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Point 5
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Correct AnswerPoint 8
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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