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?