Help on Circuit Analysis

[BruceBly]

Can someone verify the values I have calculated for the circuit I have attached to this post?



R1
7.14964894 V
1.8 kΩ
3.972027 mA
28.3986 mW

R2
5.2847380778 V
3.7997383098 kΩ
1.390816 mA
7.3501 mW

R3
12.906411547 V
5.0001383958 kΩ
2.581210 mA
33.3141697 mW

R4
1.363 V
980 Ω
1.390816 mA
1.8956827 mW

R5
6.258673469 V
4.5000000000 kΩ
1.390816 mA
8.7046652 mW

R6
7.944054380 V
2 kΩ
3.972027 mA
31.554 mW

TOTAL
28.000114870 V
7.0493260819 kΩ
3.972027 mA
111.2172176 mW

 

[MrAl]

Hi,

Here are my set of calculations in the order in which they should be computed:

PR1=0.0073501
PR6=.031554
VR4=1.363
R1=1800
R4=980
R5=4500
R6=2000
IR4=VR4/R4 {directed downward}
R2=PR1/IR4^2
IR6=sqrt(PR6/R6) {directed to the left}
VR3=(IR4)*(R2+R4+R5)
R3=VR3/(IR6-IR4) *see note
VR1=IR6*R1
VR6=IR6*R6
V=VR1+VR6+VR3

Note: The current through R1 is to the right and the current through R2 is to the right so the current in R3 subtracts.

And here is the result of doing all those calculations in the same order:
PR1=0.0073501
PR6=0.031554
VR4=1.363
R1=1800
R4=980
R5=4500
R6=2000
IR4=0.00139081632653061
R2=3799.73830976834
IR6=0.00397202719023926
VR3=12.9064115471574
R3=5000.1383957503
VR1=7.14964894243067
VR6=7.94405438047852
V=28.0001148700666

and all the other values such as the power dissipation for each resistor can be found from that information.

Comparing those values to yours, you have the same values so as long as you did the other simple calculations correct that should be good.

 

[ericgibbs]

hi,
Yours answers checkout for me.

Personally I would round up some of your calculated resistance values..

eg: 3.7997383098 kΩ >> 3.8K

E.

 

[MrAl]

Hi Eric,

Wow we posted at exactly the same time

I think i would recommend rounding too. Some of the values are extremely close to whole numbers.

 

[JoeJester]

I would recommend rounding to 3 digits and scientific notation. However, I would use whatever number of digits as seen on the problem. See chart below.

 

[BeerBelly]

I guess I still got it.
I have to admit, I'm glad I don't have to do this everyday.

 

[MrAl]

Hi,

Some of us like to do it every day, well almost every day anyway
The idea is to find better and better ways to do it.

 

[JoeJester]

That table I posted was copied from Excel. What I did was fill in the known values and looked at the unknowns. When there were two parameters available, I filled in all the other data. Then using the principles of KCL and KVL, started filling in what I could until there was enough information to complete the table, one resistor at a time.

For instance, R6's values gave me the current through R1. KCL stated that the current through R3 was R6 current minus the current through R2 or R4, or R5. The voltage across R3 was the sum of the voltage drops of R2, R4, and R5.

So for me it was the simple matter of using the basic formulae, and knowing KCL and KVL.

 

[BruceBly]

Using Excel is the same way I can about getting my figures. I wish I could use Excel on tests. It makes it much easier and I would still need to know the formulas.

Thanks for everyone's help in making sure my numbers were correct.

 

[JoeJester]

I created the table on the fly, just as you would. Who is to stop you from drawing a table ... looking just like the excel ... and filling out the data.

That is how I learned to do it 40 plus years ago.