Combination Circuit Help

[dahooligans]

I was wondering if someone could help me find the values listed to the right here, I'm somewhat familiar with this combination circuit stuff but some of these resistors values are not in here and I know that in a series the current stays the same throughout that series, and parallel the voltage is the same across, so i was wondering if someone could help me find the values below and send me a completed image link I would greatly appreciate it and I would be able to figure out on my own how those worked out to be as such. Here is the link to the circuit drawing.

http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=combinationcircuit.jpg




http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=combinationcircuit.jpg

 

[infinity]

Welcome to the forum. :grin:

We're not going to do your homework for you. Post the point in the process where you're having a problem and we'll help you.

 

[480sparky]

Best way I've learned to do Ohms law problems is to list every load, and make a chart, or graph. Would look something like:


R1..........Ohms..........Amps............Watts..............Volts

R2..........Ohms..........Amps............Watts..............Volts

R3..........Ohms..........Amps............Watts..............Volts

R4..........Ohms..........Amps............Watts..............Volts

R5..........Ohms..........Amps............Watts..............Volts
etc. Then at the bottom:

RTtl........Ohms..........Amps............Watts..............Volts


Fill in the values that are given, and use the Ohms law chart to calculate the unknowns.

 

[dahooligans]

Thanks!

Thanks!

ok, what I don't understand is that the 2.5Amps between r2 and r3, that will remain the same right? voltage of r2 and r3 would each be 100volts?

parallel to that across R5 is 75volts, so would that make the voltage across R4 having to be 125Volts or just 25 Volts? I would guess 125 volts .

That's one dillema I have....the 2nd dillema is the ams....says 5 ams heading back to the positive so would that make the amperes across R4 and R5 2.5 amps also? and then R1 would be 5 amps, 100 volts?

 

[dahooligans]

Correct Answers

Correct Answers

Or on 2nd thought im guessing the voltage across 4 will be 100 - 75 = 25 volts

total voltage of the circuit = 250V? Total wattage = 1250 Watts.

 

[dahooligans]

Or on 2nd thought im guessing the voltage across 4 will be 100 - 75 = 25 volts

total voltage of the circuit = 250V? Total wattage = 1250 Watts.

 

[dahooligans]

I've updated my combination circuit using ohms law and have a chart below it to show what I've figured out so far, my only problem is I cannot figure out how much Current will be going through R4 and R5. Whether I use 2.5amps or 5amps it still doesn't balance everything out properly in the end. Do I have something wrong in the chart already? Is the total Voltage 250 or 350 volts? or something else? I'm stuck! Please help.



http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=combinationcircuit-2.jpg

 

[480sparky]

So if you've got 5 amps total on the circuit, and 2.5 amps are going through R2 & R3, how much is going through R4 & R5?

 

[dahooligans]

I appreciate the help, im a level 1 apprentice still currently in school for 10 weeks.

R4 & R5 would each have 2.5 amps also.
But what gets me or where I'm doing something wrong then is that
the total ohms of R23 and R45 would = 26.6667 would it not? 40 + 40 (80, 1/x function) + 40 (1/x function). my total seems off?

 

[dahooligans]

Unless there's not 5 amps going thorugh R6? I would assume 5 amps are going through R6 making it 10 ohms. I would be guessing my total voltage is either 250 volts?

 

[480sparky]

I appreciate the help, im a level 1 apprentice still currently in school for 10 weeks.

R4 & R5 would each have 2.5 amps also.
But what gets me or where I'm doing something wrong then is that
the total ohms of R23 and R45 would = 26.6667 would it not? 40 + 40 (80, 1/x function) + 40 (1/x function). my total seems off?

There's nothing in the drawing that states R4 & R5 are equal resistance.

But put 2.5 amps in for R4 and R5, and solve for Ohms and Watts.

Unless there's not 5 amps going thorugh R6? ....

How could there not be? That's stated on the drawing.

 

[dahooligans]

Thanks again for your help, how do these numbers look, after trying this thing for hours so many different ways this is the only way that all the numbers balance out ok. Looks good to me....you?

http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=finalcircuit.jpg

 

[480sparky]

Thanks again for your help, how do these numbers look, after trying this thing for hours so many different ways this is the only way that all the numbers balance out ok. Looks good to me....you?

http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=finalcircuit.jpg

http://s373.photobucket.com/albums/oo180/dahooligans/?action=view&current=finalcircuit.jpg

Without running some numbers, I think you missed some steps. In order to completly solve the chart, you need to add R2+R3, and add R4+R5 to get two parallel loads, (call them R7 and R8), then calculate them and call the end load R9.

 

[dahooligans]

Ya I did that just on a piece of scrap paper at home here, so I got
R7 = 80 and R8 = 80, now since they are parallel I know that R9 will be 40 ohms which gives me R1(20ohms)+R9(40ohms)+R6(10ohms)= 70 ohms (Rt)

I'm hoping that would be correct? if it is then I'm pretty sure i'm getting this stuff down pat now.

 

[dahooligans]

thanks again for the replies and help, nice to see that there are people out there who take time to help, I really appreciate it alot!!!!

 

[infinity]

thanks again for the replies and help, nice to see that there are people out there who take time to help, I really appreciate it alot!!!!

You're welcome. Now isn't that better than just giving you the answer. :grin:

 

[dahooligans]

You're welcome. Now isn't that better than just giving you the answer. :grin:

LoL it sure is, at least this way I know how I figured it all out, makes more sense and at least I won't forget it now.

Now it's on to Alternate Current and XL, Henry's, etc.. should be fun haha :S

 

[trojans4]

1) 5 amps go thru R6 so R6 is 10 ohms
2) 5 amps go thru the parallel set of R2,R3 and R4,R6. Since 2.5 amps go thru one side (R2,R3) the other 2.5 amps go thru the other side (R4,R5).
3) Since the same amount of current goes thru each side the resistance of R2 +R3 has to be the same as R4+R5.
4) R2 + R3 = 80 ohms so R4 + R5 = 80 ohms.
5) 1/80+1/80 = 2/80 =1/40 The reciprocal of 1/40 is the correct resistance of the parallel set of four resistors which is a total of 40 ohms.
6) The total voltage drop across that set of resistors (R2,R3,R4,R5) is 5x40 or 200 volts.
7) Since R5 has a 75 V drop its resistance has to be 75/2.5 = 30 ohms
8) Since R4 + R5 = 80 ohms and R5 = 30 ohms then R4 = 50 ohms
9) Voltage drop across R4 would be 2.5 x 50 = 125 volts
10) Voltage drop across R1 would be 5 x 20 = 100 volts
11) Total voltage drop across the whole circuit is 50V (R6) + 200V (R2,R3,R4,R5) + 100V (R1) = 350 volts.

I hope that helps.