Relative Power

Today - something we talked about last week in "class"

Relative power.

In the circuit below -

We have three voltage dividers, in parallel.  R1 and R4 make up the first one.

R2 and R5 are the second, and R3 and R6 the third.

As you can see, the output voltages are 4.21, 12, and 14.19 volts respectively.

How do we calculate the voltages?  With the Voltage Divider Equation:

Where R1 is the bottom resistor, and R2 the top resistor (connected to the power supply)

So in our case, for the first divider we have R1 on top and R4 on the bottom:

So Vout = (1KΩ/(1KΩ+4.7KΩ))*24 volts = 0.1754 * 24 volts = 4.21 volts

We get the other two voltages the same way.

Now how much current is flowing in each divider?

That's easy:  I = E/R, where R in this case is the total series resistance of the divider in question:

So for the first divider, I = 24V/5.7KΩ = 4.21 mA

For the second divider, I = 24V/9.4KΩ = 2.55 mA

For the third divider, I = 24V/11.5KΩ = 2.09 mA

Now for relative power:

Let's look at R1, R2, and R3:

Which one consumes the most power?

That's easy - since R1, R2, and R3 are the same value, and P=I²*R;

The resistor with the most current through it will dissipate the most power.

Since 4.21 mA is the highest current, then R1 will dissipate the most power.

The power dissipated by R1 will be: 4.21mA²*4.7KΩ = 83.30 mW

The power dissipated by R2 will be: 2.55mA²*4.7KΩ = 30.56 mW

Notice that even though the current through R2 is more than half that of R1, that the power dissipated by R2 is less than half that of R1... Why?

Now let's take a look at a simulation from the circuit simulator at http://www.falstad.com/mathphysics.html

scroll down to the "Electrodynamics" section and select 

Analog Circuit Simulator Applet

Here you can build test circuits and see them operate - (on a computer with JAVA installed)

I set the circuit up to show relative power in color - brighter Red means more power:

This agrees well with what we calculated.

I built this circuit in the shop, and I took a thermal picture of it:

And you can see that the thermal picture agrees pretty well with the calculations.

So what do we take from this?

For a given resistor, when the current doubles, the power goes up by four times.

Or... Power is proportional to current squared.

P  I²

That's it for now.

Cheers!

Kirchoff's Voltage Law

Here is a powerful concept, that may seem a little trivial at first,
but its usefulness will grow if you stay with electronics stuff:

Put simply, Kirchoff's Voltage Law states that the sum of the voltages around a closed loop will always equal zero.

So first - some rules:

1. Assign a current, I, and stick with the direction
2. Assign polarities to the components in your circuit.
1. Voltage sources are "-" where the current goes in, and "+" where the current leaves.
2. Voltage consumers (resistors) are "+" where the current goes in, and "-" where the current leaves.
3. Write your equation, starting anywhere, and using the sign of the component where the current enters it.

So in the example above:

Since we know from Ohm's Law that Voltage = Current x Resistance:

So let's take a look at the result and see if it works:

So I = 1 amp

That makes the voltage drop across R₁  = 1 amp x 8 Ω = 8 volts

Similarly, the voltage drop across R₂ = 4 volts

So now we revisit the Kirchoff Voltage Loop and see that:

-12volts + 8volts +4volts = 0

One way this is useful is to consider the case of a blown fuse.

With the fuse blown, no current flows, but with the meter leads attached, a very small amount of current flows - indicated above by the grey arrows.

And we can use the Kirchoff Voltage Law to predict what the meter will read if the fuse is blown:

If  V_m  is the meter voltage, then the new loop becomes:

Now we find the current, I, in the circuit:

So the voltage drop across R1 = 12Ω x 1.2 micro amps (uA) = 14.4 micro volts (uV)

and the voltage drop across R2 = 8Ω x 1.2 uA = 9.6 uV

So now, without using the current to calculate the voltage across the meter resistance,

We can see that 12 volts - 14.4 micro volts - 9.6 micro volts -   V_m  = 0

And so  V_m = 12 volts - 24 micro volts, which = 11.999976 volts.

Since most meters won't display 8 digits, the meter will read 12 volts.

Conclusion:  If a fuse is blown in a series circuit, the voltage across the fuse will equal the source voltage.

Cheers for now.

Some Thoughts on 1/4 Watt Resistors

Today we visit power and how it affects the limits of voltage we work with when building circuits with the standard 1/4 watt resistor:

How much voltage can I put across a 1/4 watt resistor and not exceed its power rating?

Power, P = I*E

But I don't know I and I want to know E...

E = P/I

What does I equal?

Well, with Ohm's Law, I know that I = E/R, and I can substitute!

So...E = P/(E/R)  which, when re-arranged gets; E = P*R/E

Multiplying both sides by E:

E² = P*R

Since we are looking for E, we take the square root of both sides:

E = SquareRoot(P*R)

When we plug ressitance values for up to 5KΩ resistors, we get the graph above.

How Power and Current are Affected by Resistance

In keeping with the simple voltage divider theme, let's look at a similar circuit:

Here, R1 can be varied from 0Ω to 25Ω.

What is the voltage across R1?

Use the voltage divider formula - V (across R1) = (24volts x R1)/(R1+R2+R3)

So if R1 is 0, then the volts across it are zero.

If R1 is at its maximum of 25Ω then the volts across it will be 20 volts.

Now for a new thing:

Power

Power can be computed several ways, but the easiest is just that power is current x voltage of a component.

So the power dissapated (consumed and radiated away as heat) for R1 is
(Voltage across R1) x (Current through R1)

So if R1 = 0, the power is zero.

If R1 = 25Ω what is the power?

Well we know the volts, but we need the current:  I = 24V / (4+1+25) = 0.8 amps

So the power is now just 20 volts x 0.8 amps.  Which is 16 Watts.

But the maximum power does not occur when R1 is minimum or maximum: (0 or 25Ω).

See the graph below:

Power seems to peak when R1 = 5Ω.

Why?

Link to Chapter 3 Stuff

Here's a link to Chapter 3 related stuff:

Direct Current Folder

The Meter is Part of the Circuit

In my last post, I drew a simple voltage divider circuit.

Two 6.8 Megohm resistors in series, powered by a 20 volt DC (20 VDC) power supply.
Vout was defined as the voltage across R1, or the resistor closest to the negative side of the battery.
Using the voltage divider equation, the output voltage was calculated to be 10 VDC.
The guys in the shop built this little circuit;
-and found that the actual voltage displayed was 7.45 VDC.



...and I asked, "Why?"

Well it turns out that a typical digital volt meter has about 10 Megohms of impedance.

Impedance is just resistance to current flow, but a term used when describing both AC and DC circuits for completeness - more on that later.  For now we can think of meter impedance as just the same thing as resistance.

When the meter leads are attached to the circuit above, you can see that we now have 10MΩ of resistance in parallel with R1.

So now we need to determine the equivalent resistance (Req) of R1 in parallel with the meter.

For two resistors, that's easy:

Use the product over sum rule, which is for two resistors in parallel -

Req = (R1xR2)/(R1+R2)

So Req = (6.8MΩx10MΩ)/(6.8MΩ+10MΩ)

Req = 68/16.8 MΩ  = 4.048 MΩ

So now we re-visit the voltage divider equation:

Vout = (Vin x R1)/(R1+R2)

So Vout now is:   (20V x 4.048MΩ)/(6.8MΩ+4.048MΩ)  =  7.46 volts

We got 7.45 Volts - so this was spot-on.  The difference between 7.46 and 7.45 can be attributed to the 5% tolerance resistors we were using, and really, .01/7.46 = .0013 which gives a 0.13% error.

Not bad!

Cheers for now

Voltage Dividers

Today we talked about voltage dividers -

We used a DC Power supply and some resistors to measure what we built -

Here's the voltage divider example:

The circuit is as shown.  What is the value of V-out in volts?

The standard voltage divider equation is:

                                Vout = ((V-in)*R1)/(R1+R2)

So in this example we have:

     Vout = (20V*6.8MΩ

)/(6.8M

Ω

+6.8M

Ω

) = 10 volts

But we really don’t need to memorize a formula for this-

Good old Ohm’s Law will serve us just fine.

First, we need to find the total current flowing in the circuit...

To do that, we need to find the total resistance.

The total resistance here is just R1+R2

So the total current is found using Ohm’s Law - I = E/R

I = V-in*(R1 + R2) = 20V/13.6MΩ = 1.47 uA (micro-amps)

Now the voltage out is easy now that I know the current...

V out = I*(R1)

If we substitute the equation for I with this one, we get:

V out =  V-in*R1/(R1 + R2)

Which looks an awful lot like -

Vout = ((V-in)*R1)/(R1+R2)

Which is the standard voltage divider equation for two resistors.

We used a voltmeter to measure this.

The meter had 10 M

Ω input impedance.

We expected 10 volts like we calculated, right?

But we only read about 7.5 volts. (Actually, we read 7.45 volts)

Why?

A Hot Cup of Coffee

Get Me a Cup of Coffee!

How much power does it take to heat up a cup of coffee?

Specific Heat of Water - (OK - coffee, but coffee's mostly water)

4.186 Joules/gram-Deg C

Specific Heat of the Coffee Cup - (Which weighs about half as much as the water....)

1.09 Joules/gram-Deg C

8 oz coffee weighs about 225 grams

The cup weighs about 100 grams

So - how much POWER (in Watts)  do I need to get my coffee and cup from room temperature (about 68 degrees which is 20 Deg C)

up to a piping hot, but drinkable 158 Deg F, which is 70 Deg Celsius.

in 1 minute?

Now - It turns out that 1 Joule = 1 watt second.

The change in temperature = 70 deg C - 20 Deg C = 50 Deg C

So for the coffee - it takes

4.186 Joules 1 Watt Second 1 Minute 50 Deg C x 225 grams

Watts = ----------------- X --------------------- X --------------- X ----------------------------- = 785 Watts

gram Deg C Joule 60 Seconds 1 Minute

For the cup - it takes

1.09 Joules 1 Watt Second 1 Minute 50 Deg C x 100 grams

Watts = ----------------- X --------------------- X --------------- X ----------------------------- = 91 Watts

gram Deg C Joule 60 Seconds 1 Minute

So it will take 785 + 91 = 876 Watts to get my coffee hot in one minute....

(This ignores, of course, the fact that my coffee cup is losing heat to the air and the counter it sits on all the time, so it will really take more power than this....)

Now it turns out that I can buy some Nickel Chrome wire to use for heating the coffee....

24 AWG NiCr 60

1.671Ω/Ft                    ~  200 ft roll:   $22.49

Link to buy a roll of 24 awg NiChrome wire

At 120 Volts, how much resistance to I need to produce 876 Watts?

Well... Power, P = I x E  - I know E is 120V, and I know the power is 876 Watts.

So now I re-write the equation so that:    I = P/E

But I want to know R -

I know R = E/I so I can re-write that equation  so that:  I = E/R

Now with SUBSTITUTION I get -

E/R = P/E.     Multiplying both sides by R gets me:    E = PxR/E

Multiplying now both sides by E gets me:  ExE= PxR

Dividing both sides by P gets me: ExE/P= R

So the resistance I need is:  120 volts x 120 volts / 876 Watts   = 16.4 ohms

So how many feet of this wire do I need?

16.4 ohms / 1.671 Ω/Ft = 9.81 Feet = 118 inches...

But WAIT! - That's the resistance at 20 degrees C -
What happens to the resistance of wire as it heats? It goes up.

So for a 50 degree C change in temperature, it turns out that the resistance will go up about 20%.

So at 70C I really have about 1.671*0.2 + 1.672 = 2.006 Ω/Ft

16.4 ohms / 2.006 Ohms/foot = 8.175 feet or 98.1 inches

Now if I wind this stuff in a 3/8” diameter helical coil pattern, allowing for one wire diameter width between the turns... I can get the length down to about 5 inches.

Wire Gage Dimensions

Online Helical Coil Calculator

So when I bend the coil in half, I can now fit it into the cup...

Now before I wire it  up to a plug, I’d better figure out if it’s too much current for the breaker!

Hmmm - I = E/R so the current will be about 120 volts / 16.4 Ohms  = 7.3 amps at 70C.

At room temperature, though, the current will be about 20% higher... or 7.3*1.2 = 8.8 amps.

So no worries there!  Ahhh!  A nice hot cup of coffee -

(* DANGER!!!  Don’t really do this - risk of electrocution!  this is just a thought problem! *)

That’s it for now

Parallel Resistors Worksheet - Solution

Here's a link to the solution for the parallel resistors worksheet -

Parallel Resistors Worksheet #1 Solution

Parallel Resistors Worksheet

Here's a link to a worksheet for parallel resistors...

Parallel Resistors Worksheet

Link to excellent circuit simulator

Want to understand how circuits work?

Check this out -

http://www.falstad.com/circuit/

Done with Batteries, On to Direct Current

Battery topic finished today - exams graded, and good scores.

Now on to Direct Current -

Ohm's Law will be a centerpiece of this -

And use of basic Algebra -

We discussed balancing equations, and substitution today.

Example -

A 3600W Heater on a 240VAC circuit isn't working.... When you measure the resistance of the heater elements, what resistance reading do you expect for a good element?

Well - It seems we know the voltage, and we know the power - How to get at resistance with just plain old Ohm's Law?

Well -

Let's look at what we want to get -

R = E/I  - Resistance = Electromotive Force (Voltage) over Current (I)

Power, which we know, = E x I.

Since we know the power and voltage, but not the current, we can re-write this equation in terms of what we don't know:

So if we divide both sides by E, we get:   P/E = E/E x I   ----> P/E = I  ----> I = P/E

Now for the real POWERFUL tool!  - SUBSTITUTION

So if we want to know R, and we know R = E/I, and we ALSO know that I = P/E,  we can SUBSTITUTE P/E (which we DO know) for I (which we DON'T know) in the first equation:

So now we have:

R = E/(P/E) ---->  R = ExE/P  ------->  R = 240 x 240 / 3600  = 16 Ohms.

So - the resistance reading we expect if the elements are good, is 16 Ohms.

That's all for now - Cheers.

Battery Internal Resistance

Yesterday we re-visited the topic of internal resistance. To demonstrate the idea, 6 volt lantern batteries were used. The battery voltage was measured with a volt meter; then A 1 ohm 10 watt resistor was placed at the output terminals. A second voltage reading was immediately taken.

From these two readings we can determine a rough estimate of the battery's internal resistance. Can you figure out how?

Why is internal resistance important? Well- if you've ever tried to start your car on a cold day, I'll bet you were hoping your car's battery had lots of cold cranking amps! Or put another way, low internal resistance.

Here's a good link to the concept along with some info on how those fancy battery testers work:

http://batteryuniversity.com/learn/article/how_to_measure_cca_cold_cranking_amp/

Types of Batteries

This week we finish up with batteries -

The classic lemon battery -
http://www.youtube.com/watch?v=DXir_ORHOGA  lemon battery
Different battery types -
Dry cells, wet cells, "gel" cells to name a few.
http://www.youtube.com/watch?v=ITzAZFYFo_0  Aluminum Carbon and Salt Water
http://www.youtube.com/watch?v=DFtepc9Tpmc  How are lead acid batteries made?

What makes a battery produce voltage?
http://www.youtube.com/watch?v=Lh-J1-6ks6g voltaic cell principles

Useful hints and tricks with batteries.
http://www.youtube.com/watch?v=cV6kxQYGVs0  Cool thing you didn't know about lantern batteries...
http://www.youtube.com/watch?v=6jjNHmMB5ns  Kick start a battery that won't charge

Read up on Chapter 2 in the book and follow along...

Cheers for now

Links to stuff on my Google Drive

Here's a link to my Google Drive section for the Electronics Course.  I'll put links of interesting and related stuff there.

Electronics Stuff

Link to NEETS

Here's a link to the Navy Electricity and Electronics (NEETS) series...

www.cbtrcks.com

A bit about me

I'm John -
I work at a Marine Construction and Dredging company on all sorts of cool and fun stuff. Big electrical and mechanical systems on cranes and dredges, and small data and control systems, and just about everything in-between. The gear dates from the 1930's (really) all the way to the latest satellite communications and control stuff.

I spent 20 years in US Navy submarines, mostly as a nuclear electrician.  I then took a job installing and servicing the latest high tech 3-dimensional metal printing equipment.

Anyway - I've also been programming computers since you had to do it with teletypes and punch cards, and worked with all sorts of electrical and electronic stuff.  That's good, because I LOVE this stuff!

I think it's cool, and if you follow along, maybe you will too - this is just my way of giving a little back of all the good fortune I've had in being able to be around the technology I love to work with.

So here goes!