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.

Note: 36.8 is e^-1

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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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Hint: All the other answers deal with changing the angles, but since the angle system is arbitrary (360 was just some number someone came up with) only changing the sign of the angle is acceptable.

Admittance is the reciprocal of impedance. The reciprocal of 5 is 1/5. So far so good, but what about complex numbers? Complex numbers can be expressed in rectangular, exponential, or polar format...

A basic property of reciprocals is that their product is unity:

\[5 \times \frac{1}{5} = 1\]

Similarly, exponentials:

\[5e^{j8} \times \frac{1}{5e^{j8}} = 1\]

Alternately written as: \[5e^{j8} \times \frac{1}{5}e^{-j8} = 1\]

...because \(e^{-6}\) is equal to \(\frac{1}{e^{6}}\)

Abstracting this we get that: The reciprocal of: $Ae^{jB} $ is \(\frac{1}{A} e^{-jB}\)

Polar notation is usually shown as: \[ Z = A\cos{(B)} + jA\sin{(B)}\] And in exponential form this is: \[ Z = Ae^{jB}\] Which we can easily find the reciprocal of to show the relation between impedance and admittance.

Therefore the reciprocal of Z in exponential form would be: \[ \frac{1}{Z} = \frac{1}{A}e^{-jB}\]

And in polar form since: \[ Ae^{jB} = A\cos{(B)} + jA\sin{(B)}\]

Then the reciprocal would be: \[\frac{1}{A}e^{-jB} = \frac{1}{A}\cos{(-B)} + j\frac{1}{A}\sin{(-B)} \]

So this shows that the reciprocal of impedance in polar form is the reciprocal of the magnitude and the changed sign of the angle.

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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:

• If the angle is negative then the voltage is lagging.
• If the angle is positive then the voltage is leading.

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.

This is the polar coordinates graph we used for the questions where we find impedance point when given frequency and either L or C.

The X axis is resistance and the Y axis is L-C. Draw a line from the origin through the point. The angle is smaller as the point is further from the origin, i.e. larger R. The angle is smaller as the L-C is smaller.

You should be able imagine or draw the graph and estimate the angle. No trigonometry calculation needed.

Memory tricks:

• ALL the answers to these questions are 14 degrees.
• For the test, remember: if \(X_C > X_L\), voltage lags.
• "If L is Less, it Lags!"
• "eLi the iCe man" is a pneumonic to remember that in inductors the voltage (e) leads the current (i) and in capacitors current (i) leads the voltage (e). In this circuit, the capacitor value is larger than the inductor so we know the voltage is lagging the 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.

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.

Helpful hint: In an inductor, you have to apply voltage for current to flow, so Voltage leads Current.

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Tags: arrl chapter 4 arrl module 4b

\[\text{phase angle}\:\phi=\tan ^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\]

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.

If you are a track and field person, it is race 14. XC is s Cross Country guy and XL is the big voltage shot put guy. If XC is going faster (higher number), Voltage is lagging. If XL is going faster, Voltage is leading.

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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."

Another way to look at this is that impedance is the act of preventing access or making access difficult (like a ticket taker at the theater). The opposite (inverse) of that is allowing access or admitting it (like when the ticket taker lets you in to the theater).

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