The charge and discharge time of a CR circuit is determined by the time constant. With large amounts of capacitance and high resistance the time constant can be several hours, so beware of large capacitors in electronics equipment. In an RC circuit assuming there is no initial charge on the capacitor it takes a time of R x C seconds to charge a capacitor to 63.2% of its final value.
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This is an electronics definition that you'll have to memorize. For what it's worth, the correct answer is the only one without the word "discharge" in it.
Further reading: https://en.wikipedia.org/wiki/RC_time_constant
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Susceptance is the reciprocal of reactance, just as conductance is the reciprocal of resistance. When you convert reactance to susceptance, the sign of phase angle is reversed but the magnitude of the vector remains the same.
For example: If your phase angle is \(+θ\) degrees, the reactance is \(\cos(θ)\), whereas the susceptance is the \(\cos(-θ)\), the same magnitude as the reactance.
For more information, see: http://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/susceptance-and-admittance/
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Time constant TC or \(\tau\) is given by:
\[TC \:[\text{seconds, s}] = R\:[\text{ohms}, \Omega] \times C\:[\text{farads, F}]\]
This circuit contains two \(220\ \mu\text{F}\) capacitors and two \(1\text{ M}\Omega\) resistors, all in parallel.
The first thing to remember, capacitance in parallel INCREASES and resistance in parallel DECREASES.
For resistors in parallel:
\begin{align} \frac{1}{R_{\text{total}}} &= \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \end{align}
So, keeping consistent units in \(\text{M}\Omega\): \begin{align} R_{\text{total}} &= \frac{1}{ \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}}\\ &= \frac{1}{ \frac{1}{1\text{ M}\Omega} + \frac{1}{1\text{ M}\Omega}}\\ &= \frac{1}{ \frac{2}{1\text{ M}\Omega} }\\ &= \frac{1\text{ M}\Omega}{2}\\ &= 0.5\text{ M}\Omega\\ \end{align}
For capacitors in parallel:
\begin{align} C_{\text{total}} &= C_1 + C_2 + \ldots + C_n\\ &= 220\ \mu\text{F} + 220\ \mu\text{F}\\ &= 440\ \mu\text{F}\\ \end{align}
For the time constant:
\begin{align} \tau = TC &= R_{\text{total}} \times C_{\text{total}}\\ &= 0.5\text{ M}\Omega \times 440\ \mu\text{F}\\ \end{align}
Replace the SI prefixes Mega \(\left(10^6\right)\) and \(\mu\) (micro) \(\left(10^{-6}\right)\):
\begin{align} \tau = TC &= \left(0.5 \times 10^6\ \Omega\right) \times \left(440 \times 10^{-6}\ \text{F}\right)\\ &= 0.5 \times 440\\ &= 220\text{ seconds} \end{align}
TEST TIP: For questions E5B04, E5B05, and E5B06 the correct answer is the one equal or closest to the microfarads value in the question.
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When AC passes through a component that contains a finite, nonzero susceptance, energy is alternately stored in, and released from, a magnetic field or an electric field. In the case of a magnetic field, the susceptance is inductive. In the case of an electric field, the susceptance is capacitive. Inductive susceptance is assigned negative imaginary number values, and capacitive susceptance is assigned positive imaginary number values.
Hint: Reactance is reciprocal.
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This is a definition, so just learn it:
\[ \text{admittance} = {\text{conductance}} + j {\text{ susceptance}}\]
where \(j\) is the "j operator", electrical engineering's name for the infamous \(i\), the so-called "imaginary" number \(\sqrt{-1}\).
This makes admittance a complex number, which by definition has a "real" part and an "imaginary" part. The value of susceptance is still a real number, but it's described as the imaginary part of admittance because of that \(j\) attached to it.
Complex numbers are a convenient way to describe the relationship between two components, one having a magnitude and the other having an angle or phase. This occurs everywhere in AC current due to its cyclical nature, and thanks to Euler's formula and the concept of \(i\) it's possible to do otherwise difficult calculations using relatively simple vector algebra and some trigonometry.
\[ \text{susceptance} = \frac{1}{\text{reactance}} \]
In mathematics, we describe \(\frac{1}{x}\) as the "inverse of x." So here we can say susceptance is the inverse of reactance.
Just as we use the letter \(X\) to represent reactance, we use the letter \(B\) to refer to susceptance, so we can write that electrical relationship more succinctly as:
\[ B = \frac{1}{X} \]
This relationship is akin to the relationship between conductance and resistance: conductance is the inverse of resistance.
Reactance opposes the flow of alternating current, while susceptance allows the flow.
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Given: \begin{align} X_C &= 500\ \Omega\\ X_L &= 250\ \Omega\\ R &= 1\text{ k}\Omega = 1000\ \Omega \end{align}
We can calculate the phase angle using: \[\text{phase angle }\phi = \arctan{ \Big ( \frac{X_L - X_C}{R} \Big ) }\]
Pay attention to the sign of the angle:
Plug in our given values: \begin{align} \text{phase angle }\phi &= \arctan{ \Big ( \frac{250\ \Omega - 500\ \Omega}{1000\ \Omega} \Big ) }\\ &= \arctan{ \Big ( \frac{ -250\ \Omega }{1000\ \Omega} \Big ) }\\ &= \arctan{( -0.25 )}\\ &\approx -14^\circ\\ \end{align}
WARNING: If you are using a calculator to calculate the \(\arctan{(\ldots)}\), make sure the calculator mode is set to degrees and not radians. Using the wrong mode will give you the wrong answer!
------ or ------
The total reactance \(X\) in the circuit is \(X_L + X_C = 250\ \Omega +(- 500\ \Omega) = -250\ \Omega\). (capacitive reactance is regarded as negative when we are doing calculations on imaginary numbers).
Since we now know that the circuit reactance is capacitive we can immediately say that the voltage lags the current.
To calculate phase angle we use
\[\tan{ (\text{phase angle }\phi) } = \frac{X}{R}\]
Where: \begin{align} X &= \text{total reactance} = X_L - X_C\\ R &= \text{total series resistance} \end{align}
So \begin{align} \tan{ (\text{phase angle }\phi)} &= \frac{-250\ \Omega}{1000\ \Omega}\\ &= -0.25 \end{align}
To get the phase angle we now use the inverse tangent function of a calculator.
\[\tan^{-1}(-0.25) \approx -14^{\circ}\]
Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if \(X_C > X_L\), voltage lags.
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\[\text{phase angle}\:\phi=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\]
If the phase angle is negative then the voltage is lagging.
If the phase angle is positive then the voltage is leading.
\begin{align} \text{phase angle}\:\phi&=\tan ^{-1}\left(\frac{75-100}{100}\right)\\ &=\tan ^{-1}\left(-0.25\right)\\ &=-14.0362435^{\circ}\\ &\approx-14^{\circ} \end{align}
If you are using a calculator make sure it is in degrees and not radians.
Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if \(X_C > X_L\), voltage lags.
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A helpful way to remember the relationship of current and voltage in inductors and capacitors is ELI the ICE man:
In a capacitor, symbol C, current (I) leads voltage (E), by 90 degrees. In an inductor, symbol L, voltage (E) leads current (I), by 90 degrees.
Another helpful way to remember this is, capacitor and current both start with a C. And for the inductor, it is opposite, from the capacitor.
To understand what is happening, consider an uncharged capacitor which has no voltage across it. The voltage appears as the charge flows into the capacitor. (Flowing charge is current.) So therefore, current leads voltage. Inductors act opposite as a change in voltage changes the current flow or voltage leads current.
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A helpful way to remember the relationship of current and voltage in inductors and capacitors is ELI the ICE man:
In a capacitor, symbol C, current (I) leads voltage (E), by 90 degrees. In an inductor, symbol L, voltage (E) leads current (I), by 90 degrees.
In this case, with an inductor, VOLTAGE (E) leads CURRENT (I).
Also CiViC acronym is helpful. Clv = Capacitor = Current leads Voltage or iVlC=inductor: Voltage leads Current.
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\[\text{phase angle}\:\phi=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\]
If the angle is negative, then the voltage is lagging.
If the angle is positive, then the voltage is leading.
\begin{align} \text{phase angle}\:\phi&=\tan ^{-1}\left(\frac{50-25}{100}\right)\\ &=\tan ^{-1}\left(0.25\right)\\ &=14.0362435^{\circ}\\ &\approx14^{\circ} \end{align}
If you are using a calculator, make sure it is in degrees and not radians.
Test Tip: ALL the answers to these questions are 14 degrees. For the test, remember: if \(X_C > X_L\), voltage lags.
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Per Wikipedia:
"In electrical engineering, admittance is a measure of how easily a circuit or device will allow a current to flow. It is defined as the inverse of impedance. The SI unit of admittance is the siemens (symbol S). Oliver Heaviside coined the term admittance in December 1887."
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There's no scientific "explanation" for this answer; Susceptance has merely been assigned the letter "B".
Here's a handy list of terms:
Susceptance (B) is the reciprocal of Reactance (X)
Conductance (G) is the reciprocal of Resistance (R)
Admittance (Y) is the reciprocal of Impedance (Z)
Elastance (S) is the reciprocal of Capacitance (C)
Reluctance (ℜ) is the reciprocal of Inductance (L)
Silly way to remember: B is the only one that is susceptible to being an answer. (A B C D)
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