This post has been edited. The reason is the original post was based on the logical flaw that the total current in the circuit would not change when we connected the ground clip. If you read the origial text, all calculations was correct for a current of 9A. The comment from Vtile in reply #60 is based on the original text. In the last post we saw that the current flowing from the water heater split into two paths. One path is the neutral wire, and one path is via the ground clip, and from there into the PE wire.
We have some options when we choose how to calculate the value of these two currents. Which method you choose depends on several things, among them which method
you prefer, and which data are available for the items that are part of the circuit.
When we do calculations we can almost always discuss which method is "best". If we don't try to cheat the laws of physics we will always get the same answer. Sometimes we will maybe use slightly more time on the calculations, but that is not relevant for us now.
If we look at the last drawing, we can see that we have only one resistor in the neutral wire, and that there are three resistors in the path through the PE wire. The first we will do now, is to transform those three resitors in PE into one resistor.
They are connected in series, so we can just add the values together.
I mentioned earlier that we must pay attention to how we name all our parts. Since we now will transform these resistors into a new "logical" resitor, we will give it an unique name. We will name it
RPRP = RPROBE + RPEC + RPE
RP = 0.0664 ohm + 0.0336 ohm + 0.1 ohm
RP = 0.2 ohmThis gives us this circuit to calculate
I have indicated that we don't know the value of I in this circuit. Before we connected the ground clip, it was 9A, but because we changed the circuit when we connected the clip, the total current will also change.The first thing we must do, is to find the value of this new current.
If we look at the previous drawing we can see that the resistance in RHEAT (the water heater) is important to know if we want to know the current in the circuit.
Almost all kind of heaters are made from a material that does not change its resistance when the temperature changes. This is the same as it is stable when the power or current changes. We will use this knowledge when we calculate RHEAT.
The "secret trick" is to calcultate the resistance based on the current we know the value of, namely the current we had
before we connected the ground clip. We can do this, because the resistance is stable regardless of the current.
This is the circuit without the ground clip, and I have added all known values.
To find RHEAT, we must find the voltage across it.
Kirchhoffs voltage law (KVL)
https://en.wikipedia.org/wiki/Kirchhoff's_circuit_lawstells us that the sum of all voltage drops should be the same as the source voltage.
Since we know both the resistance in RL and RN as well as the current, we can easily find the voltage across each of them.
Usually we will specify the values so we know we get it right.
U
RL = RL x I
U
RL = 0.1 X 9
URL = 0.9VU
RN = RN X I
U
RN = 0.1 X 9
URN = 0.9VWhen we add these two values to the drawing we get this.
I have also indicated the voltage across the resistors. This can be done in several ways. Most textbooks wil use an arrow. Some will use an arrow pointing down, I am used to an arrow point up. I guess some people can spend a lot of time and words on discussing pros and cons. What is most important, is that you stick to only one standard. Else you will confuse yourself. The way I see it, is that the arrow point to the higher potential. Potential is the same as voltage.
The way I see it, is that the arrow point to the higher potential. Potential is the same as voltage.
We can see from the figure that we can use Kirchhoffs voltage law to calculate the voltage across RHEAT.
U
RHEAT = U - U
RL - U
RNU
RHEAT = 230 - 0.9 -0.9
URHEAT = 228,2VThe resistance of RHEAT is determined by the voltage across it, and the current flowing through it.
R = U / I
RHEAT = U
RHEAT / I
RHEATRHEAT = 228.2 / 9
RHEAT = 25.356 ohm_______________________________________________
Since we now know the value of the resistance in RHEAT, we can continue with the circuit "modified" by the ground clip.
Because we transformed RPROBE, RPEC and RPE into one logcal resistor (RP), we have only two diferent resistors to care about in that part of the circuit.
Now, we face a problem:
How can we calculate the value of the new logical R based on RN and RP?
If we look in textbooks and formula books we often see these two formulas.
I snipped this from wikipedia
https://en.wikipedia.org/wiki/Series_and_parallel_circuitsI am a strong believer in trying to understand the secret hiding in the formula instead of just looking it up.
There is also another "secret" with formulas. Many formulas does not give new knowledge. They exist just because it makes our calculations easier. This was most important in the older days before we got pocket calculators, but are not as important today. The second formula from wiki is aspecial version only valid for two resistors. It is of the "time savings" type. Useful, but no new knowledge.
Before we continue, we will spend some time on the first formula. What does it tell us?
When we calculated the resistance of a wire, I mentioned "conductance" and "resistance".
Conductance is a value for how easy it for electrical current to flow. We measure
conductance in
siemens, and we use to say that
G = S.
Resistance is a value for how diffilcult it is for electrical current to flow. This is what we measure in
ohm, and we use to say that
R = ohm.
We can see conductance and resistance as opposites.
Think about a road with a lot of cars. If we have only one lane, the flow of cars will be limited. This is because the "conductance" of the road is low. We can also see it as the road has high "resistance".
If we have the same number of cars, but increases the number of lanes to two lanes, the flow of cars increases.
One lane will still have the same "conductance", but since we have added one lane, that lane will also have the same value for "conductance".
The new two-lane road will have a total "conductance" described as this:
G
total = G1 + G2
Back to the first formula:
It says something about 1/R
total = 1/R1 + 1/R2
What this formula actually does, is that it calculates the conductance of all resistors connected in parallel.
1/Rtotal is the same as if we wrote Gtotal =
1/R1 is the same as if we wrote G1
1/R2 is the same as if we wrote G2So, how do we tell the calculator how to do this?
Amost all calculators, except the most simple types will have one button that does the "one divided-by-something" function.
Look for a button marked
1/x or
X-1To make sure it is the right button, you can easily test it.
Enter 100, and hit the button. The answer will be 0.01. Hit the button once more, and the ansver will be 100.
And now comes something important for beginners:If we look at the first formula, we see that it calculates the conductance. We are more interested in the resistance. This means we must hit the
1/X - button once more after we push the
= button, else we will get the wrong answer.
Now we can continue with our calculations.
We know that the new R must be calculated form the values of
RN and
RPIf we use the formula we will write it as this:
1/R = 1/ RN + 1/RP
1/R = 1/0.1 + 1/0.2
When we use a calculator we will enter this
0.1
[1/X -button] + 0.2
[1/X -button] = [1/X -button] = If we did it right, we will get
0.066666667 as the answer.
This is the same as the combined value of RN and RP is
0.0667 ohmWe name this restor just R.
We will make a new drawing to help us.
Since we now know the value of all of the resistors, and the total voltage, we can calculate the value of the current.
We will name the total resitance RTOTAL.
RTOTAL = RL + RHEAT + R
RTOTAL = 0.1 + 25.335 + 0.0667
RTOTAL = 25.502 ohmNext ting now, is to calculate the current. We remember that the current must change from the initial 9A because we changed the circuit when we connected the ground clip.
I = U / R
I = U / RTOTAL
I = U / 25.502
I = 9.019AThe total current in the circuit changed from 9A to 9.019A because of the ground clip.
Does it make sense? Yes, the ground clip made it slightly easier for the curent to flow because it added a second wire in parallel to the neutral wire form the workbench / water heater.
Since we now know the value of this logical resitor (R) and the current flowing through it, we can calculate the voltage across it.
We will use ohm's law.
U = R x I
U
R = R x I
RU
R = 0.0667 x 9.019 =
0.602This means that the voltage across the resistor is
0.602VOK,
now we know the voltage across the logical resistor R. We transformed RN and RP into this resistor only because we needed to know the voltage across RN and RP.
We know the voltage now, so let's draw a new drawing where we show where it belongs.
If we look at this last drawing, we can easily see that we know the value of RN and the voltage across it. We can also see that we know the value of RP and the voltage across it as well.
We can also see that we want to know the value of the ciurent
IN and the current
IP.
Back to ohm's law:
U = R x I
I = U / R
When we want to calculate the current IN, we must use the correct values for the resistor that determines the value of IN. That resistor is RN
I
N = U
RN / RN
I
N = 0.602 / 0.1
IN = 6.02AThen we will calculate IP
I
P = U
RP / RP
I
P = 0.602 / 0.2
IP = 3,01ANow, we can check our calculations.
We know that the current flowing from the water heater into the combination of RN and RP is 9.019A.
Kirchhoffs current law (KCL) tells us that the sum of all currents flowing into a node is equal to the currents flowing out of that node
https://en.wikipedia.org/wiki/Kirchhoff's_circuit_lawsSince 3.01A + 6.02A = 9.03A, we can assume that we got it right. I mentioned something about precision and number of decimals in a previous post. I have already done the calculations with more precision than what we can account for, so we must assume this value to be correct.
It does always make sense to do some simple checks like this. Sometimes we enter the wrong numbers, or forget a decimal point or something else that will give us the wrong answer.
This is what we know about the circuit at the moment