# Technology self-tutoring: the tutor checks the importance of unplugging an unused phone charger.

I’ve read in more than one place to unplug a phone charger when it’s not being used. Doing so I find inconvenient, but I’ve done it anyway because I’ve read I should. However, I’ve doubted there was a good reason.

Today I decided to research the matter.

howtogeek considers the potential power a plugged-in, but idle (and post 2012) phone or laptop charger will draw. However, the conclusion is that it’s negligible – barely worth the trouble of unplugging the charger after use.

itstillworks.com examines fire and short-circuit risk of a plugged-in, yet idle charger. The recommendation, short of unplugging the charger, is to use it in a power strip, in a well-ventilated, cool location, away from flammable materials. Then, even if left plugged while not charging, the risk of fire or short circuit is reduced.

Should you unplug the charger when not in use? You be the judge.

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 gives a nontechnical explanation of how a metal detector functions.

Watching Curse of Oak Island on the History channel, you see them use metal detectors.

How does a metal detector work? A basic explanation is this:

1. The search coil has electricity passing through it so that it emits a magnetic field.
2. A characteristic of metals (as opposed to nonmetals) is that their electrons are moved by a magnetic field.
3. As the metal object’s electrons move, they change the electromagnetic environment. It’s a kind of “rebound” effect.
4. The detector has a detection coil that senses the electromagnetic rebound broadcast from the movement of the metal’s electrons.

Source:

electronics.howstuffworks.com

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

# The tutor tells of a purchase he made.

A few weeks back, one of my kids wanted to borrow a flash drive so he could load his project on it to take to school. We didn’t have any at our fingertips convenient to lend; each that I knew of already contained files we didn’t want to lose. My wife eventually found him one that had been given to her for a purpose served many years ago. I resolved, however, to buy each of the kids one of their own.

Surprisingly, it’s been years since I bought a flash drive. Lots of files, nowadays, are simply uploaded to email, or some other server, then downloaded elsewhere as needed. I seldom use computers outside my home, so hardly ever use a flash drive. I was expecting to buy a couple of 2GB drives, paying around 1\$ per GB.

At Staples today, the smallest flash drive I noticed was 8GB. I figured that was much more than a student would need, so began looking at the 8GB options. However, a PNY box caught my eye: two 16GB drives for \$10.97. Was this too good to be true? I asked the clerk; she said it wasn’t the most recognized brand of flash drive, so might be less expensive just for that reason. What she said seemed true to me; I’d never heard of PNY. Turning the box over, I read “ASSEMBLED IN THE USA”. I decided to buy it.

At home, I opened the box and removed the two small flash drives. From my phone, via the desktop computer, I loaded a photo onto each one to see if they both worked, which they did. The load time seems fast to me.

It turns out PNY is a consumer electronics firm based in Parsippany, New Jersey.

I’ll be sharing more consumer electronics discoveries:)

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

# The tutor offers a hint about using the Nexus 4 onboard calculator for sin, cos, and tan.

One of my New Year’s resolutions is to get acquainted with my mobile phone. Believe it or not, I’ve been lounging on the couch, panning through its menus to discover how to use its features.

My phone is a Nexus 4, running Android 5.1.1. Last night, among the apps, I discovered its onboard calculator. I love the simple design. At the right is a green tab that you can “pull out”, revealing scientific functions.

On mine, anyway, sin, cos, and tan expect the angle to be in rads. For instance, sin30° is 0.5. However, if I enter sin30 on the Nexus 4 calculator, it replies -0.9880…., which is sin(30rads).

If you’re in the same situation, you can get the answer for degrees by entering, for example,

sin((angle in degrees)xπ÷180)

You’ll probably see the π button on the green pull-out, perhaps below sin, cos, and tan.

Therefore, if you want tan45°, just key in

tan(45xπ÷180)

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

# The tutor explains his recent understanding of a function he’s wondered about.

I’ve noticed the ENG function on more than one calculator, but have never used it. I’ve always assumed it means “engineering”; since I’m not one, it makes sense that I’m unfamiliar with it.

Yesterday my curiosity finally focused on this mysterious ENG function. You access it by SHIFT ÷ on the Casio fx-260solar. If you’re in COMP mode (I haven’t tried it with other modes), it seems to change the entered number to the highest power of 103 for which the number will be > 1. Examples:

0.056 SHIFT ÷ gives 56×10-3

0.000362 SHIFT ÷ gives 362×10-6

12037059.1 SHIFT ÷ gives 12.0370591×106

While I’m not an engineer, this notation is familiar to me. I know that in electronics, it’s common to refer to 0.056A as 56mA. Similarly, 0.000178A will likely by referred to as 178µA, also known as 178×10-6A. 12400000Ω would likely be referred to as 12.4MΩ (M=Mega=106).

I have more to say about the ENG function:)

Source:

Casio fx-260solar operation manual. London: Casio Electronics Co., Ltd.

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

# From time to time, a tutor might get asked questions about electric circuits.  In the context of tutoring or just for general interest, the maximum power transfer theorem  is nice.

I last studied electronics about twelve years ago.  Ideas from it return to mind now and then.  Researching for my Nov 10 article on auto batteries, I read a remark that whatever the battery’s resistance was, the starter should match it to receive maximum possible power. I recognized the idea as a case of the maximum power transfer theorem:

Whatever the resistance of the surrounding circuit, the load resistor should match it in order to receive maximum power.

Here’s a proof using calculus:

Let’s imagine a series circuit with peripheral resistance \$R\$ and load resistance aR, where a≥0. Since the two resistances are in series, Rtotal=R+aR.

Now, since V=IR, we have I=V/R. In particular,

I=V/(R+aR)

Furthermore, the power dissipated by a resistor is given by P=I^2R. Therefore, the power in the load resistor of our circuit is

P=(V/(R+aR))^2(aR)=aRV^2/(R+aR)^2

To find the value of a that gives the maximum value of P, we take the derivative dP/da, set it equal to zero, and solve for a.

To take the derivative dP/da, we use the quotient rule:

dP/da=((R + aR)^2(RV^2) – aRV^2(2(R + aR)R))/(R + aR)^4

Set dP/da to zero and solve for a:

0=((R + aR)^2(RV^2) – aRV^2(2(R + aR)R))/(R + aR)^4

Multiply both sides by (R + aR)^4

0=(R + aR)^2(RV^2) – 2aV^2R^2(R + aR)

Factor R + aR:

0=(R +aR)[(R + aR)RV^2 – 2aV^2R^2]

Divide out R + aR from both sides:

0=(R + aR)RV^2 – 2aV^2R^2

Factor out RV^2:

0=RV^2[(R + aR) – 2aR]

Divide out RV^2:

0=R + aR – 2aR

Factor out R:

0=R(1 + a -2a)

Divide out R:

0=1 + a -2a

Simplify:

0=1 -a

Finally,

a=1

Let’s recall that, in our circuit, the peripheral resistance is R, while the load resistance is aR. We now find that for maximum power, a=1. It follows that the load resistance should be 1R=R, the same as the peripheral resistance, for maximum power.

The maximum power transfer theorem, while many never encounter it, is a fundamental part of everyday life for many others. Anticipating what we may need to know in the future is often a challenge….

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

# Tutoring science, you are naturally interested in technology.  The tutor begins an exploration of electronics.

A transistor is a device that can be used to control current flow.  Perhaps its most recognizable application is in amplifier circuits.  In that context, very small current fluctuations received by the transistor lead to very large fluctuations in its output.

A transistor’s ability to amplify current arises of its chemical composition.  It contains three “layers”:  the collector, base, and emitter.

The collector and emitter are normally rich in current carrying species, while the base is poor.  With no current to the base, it essentially “insulates” the collector from the emitter.  No current can flow across the transistor in that case.

When the base is supplied with current, it becomes conductive between the collector and the emitter.  Now, current can flow from the collector, through the base, and out the emitter.  The circuit becomes “live”.

The conductivity of the base can be described as its “gain”; this term means the amount of output current that can be expected relative to current entering the base.  The gain is typically from 50 to 200, or even more, depending on the transistor – of which there are thousands of models.  Obviously, with a gain of 150, you’ll get 150 mA out the emitter for 1 mA in the base.  This example illustrates the amplifying ability of a transistor.  Of course, they can be used in sequence to achieve any gain required.

While the explanation given above is a simplification, it is essentially correct.  I have left out some terms and complications, for now, that I will discuss in coming posts:)

Source: Wikipedia

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