Hopefully, i will be able to give you some working skills for basic electronics. I will go in depth with Ohm's Law. If your totaly new to this, i want you to take a pencil an pad, and copy out the image below.

Below is "Ohm's Triangle". It will be you 'secret decoder ring' for when you stop and want to question something i say or present.

So how does it work? Well simply cover the symbol of what your looking for and the equation will be presented!

In principal it works like so:
    If you want to calculate:
    [*]Voltage: (Cover 'V') and it shows "I * R". Hence V=I*R.
    [*]Current: (Cover 'I') and it shows "V / R". Hence I=V/R.
    [*]Resistance: (Cover 'R') and it shows "V / I". Hence R=V/I.

    You might think this is pretty basic. But you'd be surprised how many times it will keep your head on straight ;)
Posted on 2003-04-08 21:26:17 by NaN
Lets start with the "source" ;)

Voltage is what drives and powers any electrical circuit. Its the 'energy' that does things for us. Every item on an electrical circuit has one of two forms of voltage. A voltage RISE or a VOLTAGE drop.

As it sounds, a RISE is the regaining of energy in the circuit, and a voltage drop is the release of energy in a circuit. All things either fit in one of these two scenarios. Even the WIRE itself! We tend to ignore this tho, because you need MILES of wire to have any observable potential difference (energy difference) across the wire (from end to end). Circuit boards are small in comparison. So the Voltage Drop (lost energy into the wire) is said to be 0V (Zero!).

We say potential difference, or voltage drop, because everything in the natural world requires a price. If an electron bumps into one atom, some energy is lost! (However, very little ;) ). The electron had more energy before it hit the wayward atom, and is now left with a little less. Hence the voltage drop across the Atom would be:

V(atom) = V(before the bump) - V(after the bump) = 0.0000000000000000000001V.

Im exagerating this a bit, but you should be getting the point.

This brings me to another point. Conductors vs. Insulators. Things like copper wire is a good conductor because it has a unique atomic arrangement that allows electrical current to flow through it with minimal bumping into its atoms along the way! Hence to get through a conductor, there is almost no lost energy (Voltage Drop).

Insulators on the other hand, try their best to give every electron a run for its money. Its like Bill Gates trying to get through an entire football team, and sacking the quaterback. With this much "friction" all the energy is lost between the insulator, and no Voltage is found on the other side. With no voltage, there is no insentive for current to flow. So insulators protect and guide the flow of electrons (current) down conductors, which are far easier to travel! There is an important point here:
Electrons, always seeking a way to ground, will predominantly take the path of least resistance to get there!


Voltage levels on a wire can be seen as water levels in a trench. The level of water is the same at all points in the trench. If the trench has four taps into different directions, the water level is the same value at each end.

The same thing is true with Voltage levels on conductors (since we assume 0 V drop across them). If a wire has four other wires tapping of it, the Voltage at all points of the wire will be the same (if a Battery is applied).

Remember this!
Voltages in parallel are equal!:
Vt = V1 = V2 = V3 = ... = Vn
where V1||V2||V3||...||Vn


If i were to place various sized pipes on each tap in the channel, and allowed water to drain to the ground below, the overall water level at all points of the channel would remain the same, as it drained, even though each tap may be draining at a different rate (Current)! Likewise, for wires connected in parallel, the voltage at their common end is the same, even though through resistance (pipes tapping the channel), the resulting electrical currents may be different through the end of each wire.

So where does the "water" come from. Batteries, and power supplies!. They source the energy (ie have stored Coulombs of energy, wanting to return to their "happy place", a.k.a. Ground potential). You have to make a complete path to get these Coulombs interested! You cant fool them and give em a long wire with no destination at the other end! They wont move unless they are sure to get to ground!

Below is a simple circuit example. There is only one parallel "tap" to one resistor. But the voltage at the batter is the same as the resistor for that same conductor. Likewise the potential on the Ground is also the same from end to end.

The Battery creates a Voltage RISE in potential of 12V. Seing a direct path through a resistor to ground, electrical current is produced (rate of flow of Coulombs) as it heads to the Ground Potential, where they like to be!

Just how much current is not immediately determinable. We have two definite knowns: Voltage and Resistance (12V and 2 Ohms). Hence from our "Ohms Triangle" you can easily see that the current through the resistor must amount to 6 Amps (12V / 2 Ohms).
Posted on 2003-04-08 22:51:20 by NaN
Before i get into more examples, i should stop and explain some notation. On the above diagram you can see that there is a complete path for current to flow from the batterie's positive end (where the Coulombs are stored) to the negative end (ground). The current is shown in blue lines.

As well as current, I also show where there is a Voltage Rise, and a Voltage Drop. The battery gives a Voltage Rise, by grabbing Coulombs as they return to ground, and putting them back to where they started from. This action keeps current flowing steadily in the shown loop. Alternately, due to the resistance in the resistor, the Coulombs lose their energy into it before they reach ground. This is shown as a voltage drop. Above, each potential difference (change in voltage) are shown as curved red arrows.

The arrow head is always the more positive (or larger potential) than its tail. This holds true in the above diagram, since before current entered the resistor there was 12 Joules/Coulomb, or 12 Volts. After all the resisting and the Coulombs emerge at the other end there is 0 Joules/Coulomb left, or 0 Volts (ground potential). So again, we say there is a 12 Volt drop across the resistor.

At the Battery now, you see that Coulombs, having no more energy, can eazily make it to "their happy place" at the negative end of the battery. As they do, the battery grabs them and rips them away again, and more 12 more Joules/Coulomb is invested into them and are placed at the positive terminal. This is a 12 Volt rise.

If your a keener, you might have noticed that there is something to how the current flows, and the direction of the voltage rise/drop. Well there is. Its a very significant relationship to learn! Pay close attention here, and remember this!
If both the Direction of Current, and the Voltage arrow are pointing in the same direction, then you can be sure that the device is SOURCING (or providing) energy into a circuit.
If the Direction of Current, and the Voltage arrow are pointing in oposite directions, then you can be sure that the device is DRAINING (or consuming) energy from the circuit.


If your new, and learning/practicing at home. I recomend strongly that you follow this notation style for all circuit elements. It will also aid you in other areas not yet discussed (such as evaluating power consumption within a circuit).

Another good practice to develope as you learn, is to show the direction and value of current passing through a resistor, as well as the voltage drop arrow and its amount. If you do, at every resistor you should have all three "Ohm's Law" quantitys (Resistance, Current, and Voltage). For complex circuits, this will save you time in analysing them, trust me ;)
Posted on 2003-04-09 18:32:24 by NaN
NaN,
If after passing through the resistor, each columb has 'no' energy left then how do they cover the path from the resistor to the battery?
After all for covering that path also you need some joules of energy, dont you?:confused:
Posted on 2003-04-12 09:27:43 by clippy
Circuits are Closed Networks. They are perfectly balanced from start to finish, and everything is perfectly in balance. To this end, our assumption that wire has a 0 Volt drop across it is not entirely true. It has a very finite, and small voltage drop across them as well. It is this last bit of energy that gets them through the wire. Conversely, the wire has very little friction to oppose travel, so it doesn't require much energy to do so! (This is the perfect balance in eluding to).

:NaN:
Posted on 2003-04-12 09:42:19 by NaN
I have up to now mentioned that Voltages in parallel are equal. Conversely I should finish this point and show that:
Voltages in Series are Additive. That is they are Summed up by both their Sign and Value!

V(total) = V1 + V2 + V3 + ... + Vn


This is pretty e-z to prove in example. Anyone who has ever put batteries in you walkman might have noticed their arangement. They are staggered so through small connections, the flow of current through them are in a "S" like pattern.

If you have three 1.5 V "AA" batteries, in this "S" pattern, then the Positive terminal touches the Negative Terminal of the next one, and so on. The resulting configuration has only one Positive (at the end of battery 3) and one negative (at the end of battery 1) that will act as a 4.5V source for your walkman (1.5 + 1.5 + 1.5 = 4.5).

Say you were to flip the middle one:


[-]|||||||+) (+|||||||[-] [-]|||||||+)

The crude series circuit above has the same postive and negative terminals at batteries 1 and 3, but the voltage is at these end points are now 1.5 Volts.

This is becuase the middle battery has and inverse sign ( +1.5 + (-1.5) + 1.5 = 1.5V ).
Posted on 2003-04-12 09:55:39 by NaN
Current flow as you might have noticed above is much like water.

If you measure the flow of water in a garden hose, the flow of water at any point should be the same as the last point. This is because there is no other route to go, and there is always more water coming from the water source.

In circuits this property is the same
Currents in Series are Equal!

I(total) = I1 = I2 = I3 = ... = In


This means in the simple circuit above, the current from throught the WIRES are the same amount of current flow throught the RESISTOR as is the BATTERY itself. All three of these components are in a series loop!. However, as i showed above, its the ratio of the VOLTAGE and RESITANCE that determins how much current flows! (the ratio of "electrostatic vacuum pressure" to "wire friction" spells out how many Coulombs of electrons will move per second (V/R=I) ).
Posted on 2003-04-12 10:07:25 by NaN
Now say you had one of those tap expanders for your garden hoses. You screw on this plastic attachment, and volia you can now put three garden hoses on the same outdoor faucet. Assuming the same brand of garden hose, there is now THREE parrallel paths for the current flow the leaves the faucet. This means that while all three hoses see the same pressure (voltage) their resistance will determin the current in each.

If all three hoses are the same type, it should be safe to assume they have the same resistance. Hence the current in each hose would be V (tap) / R (hose) = I (water in each hose).

However, since this tap expansion device has rubber seas and is water tight, there is no leaking! This means the faucet must SOURCE or supply 3 * I current, such that it divides into each hose as required.

For parrallel electrical circuit branches, the analogy is the same:
Currents in Parrallel must sum up to the current that supplied each branch circuit!

I(source) = I1 + I2 + I3 + ... + In


Below is another example for three parallel branch circuits:
Posted on 2003-04-12 10:50:48 by NaN
Here is the Analysis of the above circuit. If your doubtfull use your Ohms' Law Triangle for yourself ;)

The 12 V battery is a rise or Source to the circuit. It will produce a current flow OUT of its Positive terminal (Its a SOURCE).

    [*]Nodes (A) and (B) are circuited in PARALLEL to the battery's 12V terminal.
    [*]Likewise, NODES (C) and (D) are circuited in PARALLEL to the battery's 0V terminal. (Its zeros because we show ground here ~ batteries are only marked as their POTENTIAL difference from the positive end to the negative end, its how you use them that counts ;) ).
    [*]The potential difference, or voltage between Nodes (A) and (C) are then 12V - 0V = 12V. Hence the Voltage DROPED across the resistor between Nodes (A) and (C) is 12V.
    [*]The potential difference between Nodes (B) and (D) are also 12V, because of the parallel arrangement with the battery as well. Likewise, we can know that the remaining resistors also experience a 12V drop across them!

    So now we know all the voltage values through the circuit. All elements are connected in parallel to the battery, so they all get a value of 12 Volts!

    What about current then!

      [*]Between Nodes (A) and (C) we know the Resistance is 2 Ohms, and the Voltage dropped across it is 12 Volts. Hence there must be 12V/2Ohms = 6 Amps down this Branch of the circuit. Remember Resistors drain energy from the circuit, so the current flow must be against the Voltage Vector! (Down). Likewise, Batteries Source energy, so the current flow is With the Voltage Vector!.
      [*]Realizing all voltages in the circuit are parallel and hence the same, as well as that all resistances are the same too, we can concude that there will be 6 Amps though all the resistors! (Ohm's Law here!).
      [*]Now we have both series and parrallel currents. We know the parrallel currents through all parallel resistors, they coincedently all happend to be 6 Amps!. But we dont know the series current through the battery yet! This is because we have to sum up the parallel currents to find its EQUIVALENT series values!
      [*]At node (B) we have two currents of 6 Amps leaving the node to the resistors. Hence there MUST be 12 Amps supplying this node! (Currents in parrallel rule). Likewise, at node (D) we have two currents of 6 Amps returning to it, Hence there must also be 12 Amps leaving this node!
      [*]At node (A) we have 6 Amps leaving to the resistor below it, as well as the determined 12 Amps leaving to node (B). Hence we know that 18 Amps must be supplying this node! Likewise at node (C) there is 6 Amps returning from the resistor, and 12 Amps returning from Node (D). Hence its returning 18 Amps back to the battery.
      [*]Now we have our Series equivalent current through the battery. It MUST be 18 Amps, since current in series with Node (A) and Node (C) to the battery must be equal!

      The entire circuit is analyized!. The battery sources 12 Volts and 18 Amps of current results thought it from the resistive LOADING of the circuit elements.

      A little more quick math with Ohms Law and you can deduce that the entire circuit must be equivalent to (12 V / 18 A = 2/3 Ohm = 0.666667 Ohms ) Resistance to the battery!.

      So now, if you had to, can substitute the entire above circuit of restors as one resistor of 2/3 Ohms in series with a 12 V battery, since current wise to the battery, this IS the same amount of resistance the parallel circuit offers it! This latter point is an early introduction to circuit reduction ;) . Another good skill to be had in circuit analysis and design.
Posted on 2003-04-12 11:13:17 by NaN
But what if you didnt know what the battery source was. You only new that you have to provide 6 Amps of current into three parallel resistors (for some reason) of 4 Ohms each. (Note the resistance here has changed for this example!)

Well you need to reduce the circuit into a series equivant resistor. From above a similar analysis should show that if each resistor needed 6 Amps, then you need to supply 18 Amps for all three 4 Ohms resistors in parallel to have 6 Amps each. (In this case we didnt use the parallel voltages to determine each resistors current, it was given from the start as a design requirement!). From Ohms Law, if we knew the equivalent resistance, we could multiply it by 18 Amps to find out the Equivalent Voltage Drop across the entire equivalent network. Since a battery would be involved to source this series current, then it would be in parrallel with the equivalent resistor, and hense be required to Source the same Voltage value, and the problem would be solved!

So the stumping point is How to determine the resitance of the network ahead of time?

Well resistors in series are easy to realize. They effectively extend the resitance coulombs will experience, and further hinder the travell. So:
Resistors in Series are additive. That is they can be summed up to an equivalent resistance.

R(total) = R1 + R2 + R3 + ... + Rn


The problem is, we have no series resistors. They are all in parallel for our problem. So how do we treat them?

Well i could tell you, but it would be better ohms law practice to prove to you these equations:
Posted on 2003-04-12 11:52:40 by NaN
So now we have it!

The series equivalent Resitor for our network must be:

1/Req = 1/4 + 1/4 + 1/4 = 3/4

Thus Req = 4/3 or 1.333 Ohms

Our Voltage Drop across this equivalent Network would be:

Veq = 18 Amps * 1.333 Ohms = 18 * 4/3 = 6 * 4 = 24 Volts!


A quick double check at any resistor: Ir = Vr / Rr = 24 V / 4 Ohms = 6 Amps!

Ok! Its now Solved! We would require a 24 V source to produce 6 Amps down three parrallel 4 Ohm resistors!

As well, we have also picked up new resistor evaluation techniques too! As a home work assignment, prove to yourself that:

For two parallel resistors only, they can also be solved by this equation:

Req = R1 * R2 / (R1 + R2)


Its True! Applying it to any two 4 ohm resistors:

Req = 4 * 4 / (4 + 4) = 16/8 = 2 Ohms

Now again, the third left over resitor and this equivalent parallel:

Req = 4 * 2 / (4 + 2) = 8/6 = 4/3 = 1.3333 Ohms!

Again the same result! Wow! :grin:
Posted on 2003-04-12 12:16:20 by NaN
Here is a Quiz for you :grin:
Posted on 2003-04-12 14:35:03 by NaN













A Waste of Space, as the Answers are to follow.... :grin: (dont cheat!)










Posted on 2003-04-12 14:36:41 by NaN
Question #1.
Posted on 2003-04-12 14:38:04 by NaN
I didnt exand 2B as must as detailled at the first two. I only give you the answers here, you should prove to yourself that im to be trusted or not ;)
Posted on 2003-04-12 14:39:52 by NaN
You should *hopefully* now see that:

Each branch circuit currents add up to the source: 10 A = 2.5 A + 2.5 A + 5.0 A

Each parallel branch sums up to 25 Volts to equal the source:

Branch 1 = 25V
Branch 2 = 5 V + 20 V = 25 V
Branch 3 = 5 V + 15 V + 5 V = 25 V


Well this is it for my Ohms Law discussions. Hope you learned something!
:alright:
NaN
Posted on 2003-04-12 14:45:17 by NaN
You might have noticed that there is alot of examples of two like resistors in parallel, resulting in half the former resistance. This is a unique property of parallel resistances. If two resistors in parallel are identical, the result is always half the starting value.

However, if two resistors are not alike and in parallel, the result will ALWAYS be less than the smallest resistor between them!

Take a look a the graph below. Here im graphing R(parallel) = 10 Ohms in Parallel to X Resistance. Where X varies from 0.01 Ohms to 300 Ohms. See how even at 300 Ohms, the result is still less than 10 Ohms (9 Ohms or So). Only at Infinity will the result be 10 Ohms!
Posted on 2003-04-12 16:53:15 by NaN