Please can anyone help me with the following :
http://www.staff.ncl.ac.uk/alex.yako...se/ladder.html
It has to work in Cygwin Bash Shell
Thanks !
Printable View
Please can anyone help me with the following :
http://www.staff.ncl.ac.uk/alex.yako...se/ladder.html
It has to work in Cygwin Bash Shell
Thanks !
So what bit are you stuck on?
- parsing the input data
- doing the calculations
I can't finish part 1
-doing the calculations
This might be useful.
So post your best effort to date, and ask a more specific question about something you're stuck on, or what to do next.
I have done this so far ( input the readings)...
---------------------------------------------------------------
----------------------------------------------------------Code:int main()
{
const int MAXSIZE=10; // absolute max size of the array
double readings[MAXSIZE]; // array declaration
int noOfRes; // actual number of readings
int carryon = 1; // initialise flag to continue calculations
while (carryon)
{
cout << "Enter actual number of the resistors. (0<n<="
<< MAXSIZE << "): ";
cin >> noOfRes; cout << endl;
// enter actual readings from keyboard
cout << "Enter the actual value of the resistors (Ohms) , one at a time, following the prompt\n";
for (int i = 0; i < noOfRes; i++)
{
cout << "reading(" << i+1 << ") = ";
cin >> readings[i];
}
cout << endl;
How i will do that ?
Calculate the overall equivalent resistance of the circuit
Why are you asking for temperarures, when the link you posted is referring to a resistor ladder?
--
Mats
Sorry about that ! ...Quote:
Originally Posted by matsp
I made the appropriate corrections
So, what about actually TRYING to write some calculations? After all, the formulas are defined in the project specification.
--
Mats
Mate im not good as you in C++, i can't find how i'll make the programm distinguish between odd and even numbers. I think i'll use if the number
of the resistor is %2=0 then do the calculations which are specified in the problem.
I want you to write me the code for that if you can ....
And I don't want to do that, because it would teach you absolutely nothing [and teach me barely anything, because although I haven't implemented this type of calcuation, I don't think it's particularly hard]. So why should I?Quote:
Originally Posted by TaBkill3r
You have the right idea of "how to check if a number is odd or not". x % 2 will be either 0 (x is even) or 1 (x is odd).
--
Mats
You really don't need to worry about odd/even. To solve this type of circuit you just have to work backwards.
The last two resistors are in series. That equivalent resistor is in parallel with the next resistor. That equivalent resistor is in series with the next resistor. That equivalent resistor is in parallel with the next resistor. Repeat until you have no more resistors.
For this problem you know the last two resistors are in series and that there are 2n resistors.
Then you've come to the wrong place. We provide help here, not free code.Quote:
Originally Posted by TaBkill3r
I think an easier way to deal with something like this is to calculate the currents in each loop. (by applying kcl and kvl laws, and a method called the "maas" currents... I'm sure there is a formal english word for this:)).
Aferwards you get for instance 5 equations with 5 unknown variables, put it in a matrix form, calculate the echelon form of this matrix.
And kabang there you have all the currents in each loop, then its just a matter of calculating the voltage which is easy since you know all currents and know all resistance values...
I think programming wise, its easier to do it this way, the only difficult part is finding the loops and all nodes and know which resistor is in which loop...afterwards its just applying some simple laws and mathematics.
I bet if you work it out like this you get even better grades!
This is an easy task.
For all series-parallel circuits do the parallel portion first. This effectively turns the parallel portion into a series portion or turns many resistors into one with a corresponding resistance value.
Once you have this then you can plug the values into Ohm's law and figure out the rest of the circuit.
Parallel: Total = 1/R1 + 1/R2 + 1/R3 + .... 1/Rn
Series: Total = R1 + R2 + R3 + ... + Rn
Ohm's law: E = IR - E = voltage, I = amperage, R = resistance
I highly recommend RIEP charts (P(ower)=I * E) to figure circuits out. You can implement this in code as well. Using RIEP charts you can figure out any circuit with little trouble.
Most of that question is a 'trick' question. Regardless of how the resistors in a circuit are arranged they always follow the fundamental parallel or series rules. If you have parallel portions, you must figure those out first.
Also remember this:
Series
Amperage is constant - hence you get voltage drops across the resistors
Parallel
Voltage is constant - hence you can figure out how many amps are flowing through each resistor in a parallel circuit
Indeed there is: "mesh" currentsQuote:
Originally Posted by GanglyLamb
Aha I'm learning each day:).
(offtopic: is there any place where one can look up jargon names of things, and translate it into any language... because thing like mesh current is not really something I find by using a normal online translation tool..).
@Bubba: indeed its an easy task, but by using the mesh current idea one can solve this in a generic way and have all currents and voltages at the same time. By using the normal approach one would first have to calculate the total resistance, calculate the current, in a parallel, divide it among the possible ways it can flow depending on the total currents in those loops etc etc... If I could use a calculator tool like maple,I would prefer to use the mesh current thing (which gives me a matrix with all solutions) then calculate everything by hand, step by step, backtracking into the whole network...
To the OP if all of the above doesnt make sense: http://www.allaboutcircuits.com/vol_1/chpt_10/3.html
Anyhow, I'm looking forward to see the solution written by the OP.:)
Thank you all for the help ! :D