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.