The charge and discharge time of an RC 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 \(R \times C\) seconds—first time constant (\(\tau\))— to charge a capacitor to 63.2% of its final value.

Note: Charge voltage values are derived from mathemetical constant \(e\). \(63.2\% \approx 1-e^{-1}\) and \(36.8\% \approx e^{-1}\).

A capacitor is typically considered fully charged after 5 time constants (\(5\tau\)).

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Tags: arrl chapter 4 arrl module 4b

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)

Hint: Why "B"? - "B"eats me!

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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}

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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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Tags: arrl chapter 4 arrl module 4b

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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Tags: arrl chapter 4 arrl module 4c

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.

\[\tan^{-1}(-0.25) \approx -14^{\circ}\]

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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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Phase angle between voltage and current through a series RLC circuit is given with the formula:

\begin{align} \phi&=\tan^{-1}\left(\frac{X_{L}-X_{C}}{R}\right)\\ X_L=100\Omega \\ X_C=300\Omega \\ R=100\Omega \\ &=\tan^{-1}\left(\frac{100-300}{100}\right)\\ &=\tan^{-1}\left(-2\right)\\ &=-63^{\circ} \end{align}

Furthermore, a negative angle indicates that Voltage is lagging Current, while a positive angle would indicate Voltage is leading Current. (Think negative is behind or lagging and positive is ahead or leading)

Therefore, The answer is: 63 degrees with the voltage lagging the current

This question is unfortunately one that is equation heavy, and it may be necessary to simply memorize the equation in order to be able to answer it in the general case. However, if a geometric approach is easier for you, consider that:

In rectangular format the described Impedance is: \begin{align} Z&=R\pm jX \Omega\\ &=100\pm j(100-300) \Omega\\ &=100- j200 \Omega \end{align}

Graphing this gives:

Which by inspection is an angle closer to \(63^{\circ}\) than it is to the alternate answer of \(27^{\circ}\), and since it is below the x-axis it is a negative angle and therefore voltage is lagging 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.

Hint: Current leads Voltage in the question.

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Tags: arrl chapter 4 arrl module 4b

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

First, you need to find the reactance (X). X is simply inductive reactance minus capacitive reactance (X = XL - XC).

Now R and X form a right triangle, R being the adjacent side and X being the opposite side. You can use the inverse tangent to get the angle from the opposite over the adjacent.

tanθ = X/R

tan−1(X/R) = θ

To determine if the voltage is leading or lagging, remember ELI the ICE man. For an inductive (L in ELI) circuit, voltage (E) leads current (I). For a capacitive (C in ICE) circuit, current (I) leads voltage (E). E in this case stands for electromotive force, which is another term for voltage.

REMEMBER: make sure your calculator is in DEGree's not RADians

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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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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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