The AC current density is strongest at the surface of a conductor, and the magnitude decreases exponentially as you get farther away from the surface. Several variables affect this distribution, with frequency being one of them. You just have to remember that the current density at the surface increases with increasing frequency, leading to a 'thinner' RF current.

The skin effect governs how far RF signals penetrate a given material.

Silly Hint: Where can you find a lot of skin on people? In their creases ("Increases" is in the right answer).

Last edited by cq.n8gmz. Register to edit

Tags: arrl chapter 4 arrl module 4e

Any wire has self inductance, which increases with the length of the wire (among other things). Since the impedance of an inductor is proportional to frequency, it is usually safe to ignore the self inductance of short wires at low frequencies. But for VHF and above a wire's self inductance may have significant inductive reactance. This reactance is often unwanted and can be minimized by keeping connections short.

Last edited by n6sjd. Register to edit

Tags: arrl chapter 4 arrl module 4e

Microstrip is an RF transmission line implemented on a PCB. Two conductor transmission lines (like coaxial cable) consist of two conductors separated by a dielectric. In a microstrip transmission line the two conductors are etched copper layers on the PCB. One is a narrow strip (trace) the other is a wide region of copper (ground plane). The dielectric is the material separating the layers of the PCB, for example FR-4.

See: https://en.wikipedia.org/wiki/Microstrip

Last edited by scoopgracie. Register to edit

Tags: arrl chapter 5 arrl module 5b

The answer is somewhat bogus as with microstrip and other high frequency designs, you use controlled lengths of connections (transmission lines) to purposely introduce phase shift which is part of tuning and matching.

In other words, short connections are not necessary other than to cut down on loss and parasitic radiation.

But if you wanted to minimize phase shift (which is rarely a design goal), then you would want short connections.

Just remember it is the only answer with "phase shift" in it.

It's also the only answer with the word "connection" in it, which is also in the question.

Last edited by cougargrl7. Register to edit

Tags: arrl chapter 4 arrl module 4e

The power factor of an RL circuit having a 30 degree phase angle between the voltage and the current is 0.866.

The power circle equation based on Ohm's law recognizes that the power factor is 1 when there is no phase angle between the voltage and the current.

Therefore taking the trigonometric cosine of the phase angle will give the power factor.

So for this question: \(\cos(30^\circ)=\frac{\sqrt{3}}{2} \approx 0.866\)

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 4 arrl module 4d

The Left-Hand Rule:

First point your thumb up, your index finger forward and your middle finger to the right. Your index finger is now pointing in the direction of the magnetic field, your middle finger is pointing in the direction of current (from - to +), and your thumb shows the direction of the force exerted.

This determines that the magnetic field is at 90 degrees to the electron flow in this question. As that is not an option, " In a direction determined by the left-hand rule" would be the only correct answer.

Last edited by wonko. Register to edit

Tags: arrl chapter 4 arrl module 4b

In a circuit having a power factor of 0.71 and apparent power of 500 VA, 355 W will be consumed.

\begin{align} \text{Power}_{\text{consumed}} &= \text{Power}_{\text{apparent}} \times \text{Power Factor}\\ &= 500 \times 0.71\\ &= 355\text{ W} \end{align}

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 4 arrl module 4d

The power factor \(\text{PF}\) multiples the power. We know that power can be calculated by multiplying the voltage \(V\) and the current \(I\).

\begin{align} \text{Power Consumed} &= V\times I \times\text{PF} \\ &= 200\:\text{V} \times 5\:\text{A} \times 0.6 \\ &= 1000\:\text{W}\times 0.6 \\ &= 600 \:\text{W} \end{align}

Note that "power consumed" refers to the power consumed by the load. The electric utility still has to generate both the real power and the reactive power. The reactive power is wasted as it is not delivered to the load.

Last edited by w0rcf. Register to edit

Tags: arrl chapter 4 arrl module 4d

The question states both ideal inductors and capacitors, so think perfect. The current just passes from the inductors (magnetic field) to the capacitors (electric field), back and forth so none of the power is lost or dissipated.

Last edited by km4wpo. Register to edit

Tags: arrl chapter 4 arrl module 4d

The true power in an AC circuit can be determined by multiplying the apparent power times the power factor. The power equation based on Ohm's law assumes that the voltage and current are in phase -- therefore, when the voltage and current are out of phase, the result of the power equation is multiplied by a coefficient known as the power factor.

Last edited by kd7bbc. Register to edit

Tags: arrl chapter 4 arrl module 4d

The power factor of an R-L circuit (a circuit having a Resistor and an Inductor), having a 60° phase angle between the voltage and the current is simply the cosine of that phase angle:

\[\cos{60^{\circ}} = 0.5\]

WARNING: Be careful that your calculator is not in radians; it must be in degrees.

Last edited by kt4obx. Register to edit

Tags: arrl chapter 4 arrl module 4d

Given:

\(E = 100 \text{ V}\)
\(I = 4 \text{ A}\)
Power Factor (\(\text{PF}\)) \(= 0.2\)

How many watts are consumed in this circuit?

The consumption with a power factor of 1.0 would be defined by Ohm's law: \begin{align} P &= E \cdot I\\ &= 100 \text{ V} \cdot 4 \text{ A} \\ &= 400 \text{ W} \end{align}

We call this result the Apparent Power, and often refer to it with the symbol \(S\):
\[S = P_{\text{apparent}}= 400\text{ W}\]

To determine the real power, the apparent power needs to be multiplied by the power factor:
\begin{align} P_{\text{Real}}&= S \cdot \text{PF}\\ &= 400\text{ W} \cdot 0.2 = 80\text{ W} \end{align}

Last edited by kv0a. Register to edit

Tags: arrl chapter 4 arrl module 4d

Only resistance (real component of impedance) consumes power. The values for the resistor, 100 ohms, and current, 1 A, are given.

\begin{align} P_{\text{real}} &= I^2 R\\ &= (1 \text{ A})^2(100 \:\Omega)\\ &= 100 {\text{ W}} \end{align}

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 4 arrl module 4d

Capacitors resist change in voltage and inductors resist change in current each by storing energy and releasing it as voltage and current fluctuate. This is called reactance. Unlike resistance, no actual power is dissipated by reactance. In purely reactive circuits there will still be measurable voltage and current. The product of this voltage and current is called "wattless" power, measured in volt-ampere reactive (VAR).

Last edited by bapaige01@yahoo.com. Register to edit

Tags: arrl chapter 4 arrl module 4d

The power factor is defined as the ratio of active (true) power P to the absolute value of apparent power S, or

\[\text{PF} = \frac{P}{|S|}\]

which is also the ratio represented by the cosine of phase angle between the corresponding voltage and current. Therefore, the power factor in this case is

\[\text{PF} = \cos(45^{\circ}) = \frac{\sqrt{2}}{2}\approx0.707\]

Silly Hint: The number "7" as it is written in a numerical value is close to a 45-degree angle. 0.707 is the only choice with "7"'s.

Last edited by w8pla. Register to edit

Tags: arrl chapter 4 arrl module 4d