# Self-tutoring about electrical consumption: the tutor looks for the typical power rating of baseboard heaters.

Investigating energy consumption, I was compelled to inquire about the wattage of baseboard heaters.

Baseboard heaters, because of their variable length, are often discussed in terms of wattage per foot.

The ones I’ve examined range from 243 to 263 Watts per foot (800 to 867 Watts per metre). The articles I’ve read give 225 to 250 Watts per foot (743 to 825 Watts per metre), for standard ones.

Source:

homeguides.sfgate.com

www.bchydro.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Self-tutoring about wiring: the tutor inquires about where 208V supply might be found.

Looking at the baseboard heaters the other day, I noticed they have two wattage ratings: one if in a 208V system, and one if in 240V.

I assumed this house is 240V, and even confirmed so at the source panel. Yet, I reflected, 208V must exist, and even be fairly common, if the baseboard heaters mention it.

Apparently, 208V is common in commercial buildings in North America.

Source:

ctlsys.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring physics or chemistry, you might encounter Peukert’s Law, although it’s probably used more by industry.

In my March 2 post I mentioned reserve capacity and Amp*hours as two ways to measure a battery’s potential output. Numerically they are convertible backwards and forwards, but the reality is not necessarily so simple, because the speed of discharge affects the total output a battery can manage. Specifically, a higher discharge rate lessens the battery’s efficiency, so that the total output will decrease the faster the battery is discharged.

Peukert’s Law gives the equation

t=T(C/(I*T))k, where

t is the actual time the battery will deliver arbitrary current I,

T is the discharge time corresponding to the given amp*hour rating,

C is the given amp*hour rating,

k is a physical constant that depends on the type of battery (around 1.4 for lead-acid).

Because of Peukert’s Law, an amp*hour rating must be given with a specific time for which it’s valid. (My reading suggests that 20 hours might be a typical time.) Therefore, an amp*hour rating might read “120Ah over 20 hours”. Such a rating implies a discharge rate of 6A for 20 hours. How long the battery can deliver a different amperage can be calculated by Peukert’s Law.

Example: Imagine a battery rated 120Ah over 20 hours. How long can it deliver 120A?

Solution: Theoretically, a 120Ah battery can deliver 120A for 1 hour, although we already know not to expect it. Peukert’s Law gives

t=20(120/(120*20))1.4

t=20(0.05)1.4

t=0.30 hours, or about 18 minutes.

Source:

batteryuniversity.com

batteryuniversity.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# Tutoring chemistry, the distinction between cell and battery is noted.

In electrochemistry, a cell is a single unit of electrical energy production. A cell comprises an anode and cathode, plus the ingredients and the environment needed for the chemical reaction that outputs electrical energy.

A battery comprises more than one cell connected so that they work together to deliver energy to a circuit.

People have come to refer to single cells as batteries. I’d say that the button-style power sources found in calculators, watches, etc are cells. If a calculator contains two of them, those two cells constitute a battery.

The typical car battery really is one, since it contains six cells connected.

Source:

Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Giancoli, Douglas C. Physics, fifth ed. New Jersey: Prentice Hall, 1998.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor explains which terminal is the anode, which is the cathode, as well as polarity.

In terms of electrochemistry, oxidation occurs at the anode; reduction occurs at the cathode.

Perhaps less obvious is the answer to the question, “Which is positive and which is negative?”

At the cathode, reduction is occurring, so it consumes negative charge. Therefore, the cathode is positive.

At the anode, oxidation is occuring, so negative charge is being liberated. Therefore, the anode is negative.

Source:

Mortimer, Charles E. Chemistry, sixth ed. Belmont: Wadsworth, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor defines two terms relating to auto batteries.

The reserve capacity (RC) of a car battery is given in minutes. It’s the duration the battery can discharge 25A at ≥ 10.5V.

Ampere hours (Ah), on the other hand, is the product of steady discharge in Amps that is possible for a duration of hours before the battery voltage slips below 10.5V. Ah is often given in a format like 50Ah@20h, meaning that, over a 20 hour duration, the battery delivered a steady 2.5A at or above 10.5V.

Mathematically, reserve capacity can be converted to Ah as follows:

Ah ≈ RC*(5/12)

However, this equation might be more a “rule-of-thumb”, since the total amount of energy available from a battery decreases with speed of discharge. In other words, a battery that delivers 2.5A for 20h might likely not manage 50A for 1h.

Source:

www.pacificpowerbatteries.com

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor briefly explains the low-pass filter.

This explanation draws on ideas from that of a high-pass filter (see my article here).

A low-pass filter sends along low frequencies but blocks higher ones. The one we’re looking at today has a resistor and a capacitor in series. As detailed in my article on the series high-pass filter, we have total impedance Z=(R2 + Xc2)0.5, where Xc = 1/(2πfC).

As the frequency decreases, the impedance of the capacitor increases, so its share of the voltage output rises. A low-pass filter will read the voltage across the capacitor. Relative to the input voltage for the circuit, V, its output will be

Vout/V = Xc/(R2 + Xc2)0.5

At very low frequency, the impedance of the capacitor Xc = 1/(2πfC)>>R, so

Vout/V ≈ Xc/(Xc2)0.5 = Xc/Xc = 1

The critical frequency, fc, is when Vout/V = 0.707. fc happens when R=Xc:

Vout/V = Xc/(Xc2 + Xc2)0.5 = Xc/(2Xc2)0.5 = 1/20.5 = 0.707.

To find fc we set R=Xc=1/(2πfC), then arrive at f = 1/(2πRC). A series low-pass filter with capacitor 4700pF and resistor 10kΩ will have critical frequency fc = 1/(2π*1×104*4700*10-12) = 3390Hz.

Source:

www.electronics-tutorials.ws

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor gives background along with a basic explanation of how a high-pass filter works.

A high-pass filter will send along high frequency signals but block low frequencies. It can do so because the impedance, Xc, of a capacitor of capacitance C, at frequency f, is

Xc=1/(2πfC)

At very high frequency, therefore, Xc≈0

The total impedance, Z, of a resistor R and capacitor C in series is given by

Z=(R2 + Xc2)0.5

Therefore, with input voltage V, and instantaneous current i, Ohm’s Law gives

V=i(R2 + Xc2)0.5

The voltage across the resitor is

VR=iR

Therefore, the ratio of the voltage across the resistor to the source is

VR/V = iR/i(R2 + Xc2)0.5 = R/(R2 + Xc2)0.5

As mentioned earlier, for high frequency, Xc≈0, giving

VR/V ≈ R/(R2)0.5 = 1

With a high-pass filter, the maximum output voltage for a high frequency signal equals the input voltage.

Convention says that the critical frequency, fc, is that at which VR/V = 1/(2)0.5 = 0.707, which occurs when Xc = R:

VR/V = R/(R2 + Xc2)0.5 = R/(R2 + R2)0.5 = R/(2R2)0.5

VR/V = 1/20.5 = 0.707

Therefore, a high-pass filter will pass any frequency higher than the critical frequency fc, where fc is calculated from

Xc = R

1/(2πfcC) = R

1 = 2πfcCR

1/(2πCR) = fc

By that reasoning, a 10kΩ resistor, in series with a 12pF (picoFarad) capacitor, placed in series, should produce a high-pass filter with critical frequency fc = 1.3MHz. The output would be read across the resistor R.

Source:

www.electronics-tutorials.ws

Serway, Raymond A. Physics for Scientists and Engineers with modern physics. Toronto: Saunders College Publishing, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor discusses the electric field between two parallel plates, then between two point charges.

A consequence of Gauss’s Law is that, from an infinite charged plane, the electric field is constant, independent of distance, and given by

E = σ/2εο

where

σ = the charge density of the plane in N/m2

εο = 8.854187817 x 10-12, the permittivity of free space.

In a real capacitor, if the plates are much higher and broader than their separation, then at a point between them, collinear with their centres, the effect is probably comparable to two infinite planes of charge. In that case, the field, regardless of position along that centre line, is given by

Enet = E2 – E1

Now, a different premise: we imagine point P between two charged particles, q1 and q2, such that q1, P, and q2 are all collinear. In this situation the field at point P depends on its position between q1 and q2 and is given by

Enet = E2 – E1 = kq2/r22 – kq1/r12

where

k = 1/(4π*εο) = 9.0 x 109

r1 = the distance from P to q1

r2 = the distance from P to q2

Source:

Serway, Raymond A. Physics for Scientists and Engineers with modern physics, 2nd ed. Toronto: Saunders College Publishing, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

# The tutor shares some research about why he gets shocked leaving the car.

When the weather turns dry and summery, I love it. However, I face a summertime hazard: getting an electric shock when I leave the car. Sometimes the shock makes me wince as I hear the “crack”.

Last night I researched the reason for these shocks. Possibly, it’s mundane: as my leather belt and even my dry skin rub against the polyester seat, they develop positive charge (the seat, negative). When I touch the metal door, my positive charge attracts electrons through my fingers – zap! In dry weather, the process is emphasized, since there is less water to bleed away static charge.

I’ll be talking more about static electricity:)

Source:

school-for-champions

physics.org

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.