Circuit calculation problem

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
Hi everyone i have tried to figure out this paper but am stuck with a few things and i dont know if am even headed to the right direction. Could anyone be kind enough to assist me with this.

A) Firsty i need to calculate total resistance, witch resistors are in parrarel and witch ones are in series?

e) Secondly they ask me to measure UR1 witch i did, but now they ask me to calculate it but i cant figure out the equation for it? How can i calculate indiviual resistors voltage?

f) 20160921_085514.jpg 20160921_085514.jpg I also need to calculate IR2 AND IR3?

I have no clue. Please help me anyone
 

AlbertHall

Joined Jun 4, 2014
12,345
Just to get you started a bit, you have calculated the value of R3+R4+R5 but that is wrong.
R3 and R4 are in parallel so you need to calculate their parallel resistance first.
Then that calculated resistance is in series with R5, so now you can calculate the effective resistance of that combination of those three resistors.
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
Just to get you started a bit, you have calculated the value of R3+R4+R5 but that is wrong.
R3 and R4 are in parallel so you need to calculate their parallel resistance first.
Then that calculated resistance is in series with R5, so now you can calculate the effective resistance of that combination of those three resistors.

Hi

Thanks for the help. It made more sense but now i have been batteling with finding out UR1. Can you get me started with that? How can i find out its indivual resistance or current?
 

AlbertHall

Joined Jun 4, 2014
12,345
R7 feeds three parallel paths. Calculate the resistance of each of those three paths, then you calculate the resistance of those three in parallel.
That resistance is in series with R7 and so now you can calculate the total current, the voltage across R7, and so the voltage across each of the three parallel paths. Then you can calculate the voltages and currents in each path.
 

ci139

Joined Jul 11, 2016
1,898
_Draft-RMTX-xlsx-PS.png PS "!!! Exception makes a rule" -- you have to account the finite precision Floating point arithmetics calculation error which goes greater with exponential calculations so it'd be reasonable to check match against 10^(-4) e.g. 0.01% . . .
 

WBahn

Joined Mar 31, 2012
29,976
You have two values listed for the total resistance -- 97.97 Ω and 92.1 Ω. I'm guessing one of them is the calculated resistance and the other is measured? I'm further guessing that the first is the calculated and the second is the measured? You have a significant difference between them (about 6%), which indicates something is wrong in one or the other. Review your calculations since the total resistance, using the nominal values, is just under 93 Ω.
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
You have two values listed for the total resistance -- 97.97 Ω and 92.1 Ω. I'm guessing one of them is the calculated resistance and the other is measured? I'm further guessing that the first is the calculated and the second is the measured? You have a significant difference between them (about 6%), which indicates something is wrong in one or the other. Review your calculations since the total resistance, using the nominal values, is just under 93 Ω.

Sorry i needed to erase that 97.97 Ω My friend gave me instructions on how to calculate it but it was wrong, so i had to make this account to get help. With the help of this forum i calculated it to be 92,904. The other is measured and the other is calculated.
 

WBahn

Joined Mar 31, 2012
29,976
The 92.9 Ω value is correct. Don't report results to more sig figs than are warranted -- your nominal resistance values are only given to at best 1% (some are only 3%) and they almost certainly have about a 5% tolerance (the measured values are in the 1% to 2% range, which is pretty typical for 5% resistors these days), so you really aren't even justified reporting the result as anything other than 93 Ω, but including one more sig fig is common practice for things like this. Also, never forget the units -- 92,904 is not a resistance, it is just a number.
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
HI

I want to thank you for your efforts. Your calculations where awsome, but the only thing i did not understand was calculating the Ur1 part. The teacher wanted me to calculate and not measure, so i did not understand the node part? is this even possible to calculate without measuring and if so am i missing something here.? Thank you again for your effort and post
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
The 92.9 Ω value is correct. Don't report results to more sig figs than are warranted -- your nominal resistance values are only given to at best 1% (some are only 3%) and they almost certainly have about a 5% tolerance (the measured values are in the 1% to 2% range, which is pretty typical for 5% resistors these days), so you really aren't even justified reporting the result as anything other than 93 Ω, but including one more sig fig is common practice for things like this. Also, never forget the units -- 92,904 is not a resistance, it is just a number.

Thank you for your message. This forum really made an impression on me. I will be staying here. You guys are awsome and super nice!!!!
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
NOW I FIGURED UR1. How foolish of me.... it was 5,642V:65=0,086
UR1= 0,086X18=1,559V

Now am facing the last problem in this paper and that is how can i calculate Ir3? i calculated R5 to be 0,070mA but how can i get IR3 value?
 

Thread Starter

Husain Al Husaini

Joined Sep 21, 2016
11
Now i figured out the Ir3 also. R5 was 0,070mA so R5x0,070=3.92V
3.92-5,642=1,722V Witch is the voltage for R3 And R4. R3X1,722V equals around 56mA. My values are a bit diffrent but thats all good. I want to thank you guys for your help and support! Awsome people! THANKS!
 

ci139

Joined Jul 11, 2016
1,898
that type of resistor matrices can be a real head ace even with 5 resistors . . .

► the one you had is not an easy one - i was not sure at all whether i manage the results or if i mange them right :confused:
especially with some unknown resistances to be found - congratulations!
 
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