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.

Memory Aid: The question refers to conductor skin effect. The portion of the correct answer - “…current flows close to the surface,” refers to “skin” (on the surface).

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

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Remember that the skin effect changes AC current distribution in a conductor as frequency increases, concentrating it more strongly at the surface of a conductor.

Think about the case of a cylindrical copper wire. For a DC current, the current density is evenly distributed in the wire. At higher frequencies, the current is concentrated at the outer edge of the wire. If you assume the magnitude of current is the same as the DC case, the same amount of current is now flowing through a much smaller effective cross sectional area. This impedes/resists AC current flow more. It's akin to having water flowing through a cylinder, and then having force the same volume of water through a tunnel with a much smaller size.

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The keywords here are stored and electrostatic field.

Let´s exclude some options: 1- While a battery does store energy, there is no electrostatic field. 2- A transformer does not store energy. 3- An inductor stores energy but as magnetic field (https://en.wikipedia.org/wiki/Inductor)

So we are left with the capacitor: It does store energy and it does it using an electrostatic field.

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Joule is the unit of measure for energy.
Volt the unit of measure of electric potential, not energy. Coulomb is the unit of measure of electric charge. Watt is the unit of measure of power (which is not the same thing as energy).

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The most common example of this is an electromagnet; most electronics students at some point try the experiment of wrapping insulated wire many times around an iron nail. When you apply current, the nail will become a magnet -- specifically an electromagnet.

The reason that the nail becomes a magnet is that the Electric Current flowing through the wire wrapped around the nail creates a magnetic field. Since the nail has a ferrite (susceptible to magnetism) core, the nail becomes magnetized.

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Simple Right-Hand Rule:

Imagine grasping a conductor in your right hand with your fingers curled around the conductor and your thumb pointing up. If the current is flowing in the direction indicated by your thumb, the magnetic field will curl around the conductor in the direction indicated by your fingers.

Thus the correct answer is "In a circle around the conductor."

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Biot-Savart law for a sufficiently long wire is: \[B=\frac{\mu I}{2\pi r}\] where:

• \(B\) is the magnetic field
• \(\mu\) is the magnetic constant
• \(r\) is the distance from the wire
• \(I\) is the current

The direction of the magnetic field \(B\) is found with the right hand rule. The magnetic field is therefore linearly related to the current in the wire.

This section of Wikipedia is helpful : https://en.wikipedia.org/wiki/Magnetic_field#Magnetic_field_and_electric_currents

Also note the right hand rule is for movement of positive charge, and the left hand rule for electron flow. The sign of the charge carrier determines which hand to use. Be positive before you write something down and leave the negativity to the sinister.

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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.

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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.

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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}

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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}

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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).

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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.

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

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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.

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

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

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