Maths - Resistors in series and parallel: Advanced

[D Skelton]

Maths - Resistors in series and parallel: Advanced

 

[Happy Jack]

Thanks for the refresher, this is most welcome. Okay. I'll come back again when I'm at home and show my workings, but here is my attempt which I haven't confirmed myself. Just dropping it now off the cuff. All answers to two decimal places.

Resistance total (Rt)
= (1 / ((1 / (12 + 6)) + (1 / 43) + (1 / 14)) + 1.5
= (1 / (0.056 + 0.023 + 0.071)) + 1.5
= (1 / 0.15) + 1.5
= 6.66 + 1.5
= 8.16R

Current total (It)
= Vs / Rt
= 230 / 8.16
= 28.19A

Resistance @ A
= Vd / I
= 62.6 / 10.43
= 6R

Current @ B = 13.41A
Power drawn @ C = 575.42W
Voltage @ D = 187.71V

 

[Robson689]

I will work out the top parallel first.

The voltage at the 12Ω resistor would be 125.16V (10.43x12) and the resistor value of the top right would be 6Ω (62.6/10.43) giving you a total of 18Ω at the top. To work out resistance in parallel it's 1/R(total) = 1/R1+1/R2 so ((1/18=0.056) + (1/43=0.023)) = 1/0.079 = 12.66Ω.

(top) = 12.66Ω
(mid) = 14Ω
(bot) = 1.5Ω

R(total) = 1/12.66 + 1/14 = 0.079 + 0.071 = 0.150 + 1.5 = 1.65Ω

Have I done something wrong? Happy Jack got 8.16Ω.


I = V/R so the current at B would be 230/4 which is 57.5A and in order to work out the power drawn at C we need the current at that resistor also because P=IV.

Current at C would be 71.875A (230/3.2) so 71.875x230 = 16531.25W or 16.53125kW

I did that wrong because voltage changes in series so I would need to work out the voltage at the resistor but to do so I need to work out the total current.

Resistance Total (Rt) = 1.65Ω
Current Total (It) =
Resistance at A = 6Ω
Current at B = 57.5A
Power Drawn at C =
Voltage at D =

 

[D Skelton]

I will work out the top parallel first.

The voltage at the 12Ω resistor would be 125.16V (10.43x12) and the resistor value of the top right would be 6Ω (62.6/10.43) giving you a total of 18Ω at the top. To work out resistance in parallel it's 1/R(total) = 1/R1+1/R2 so ((1/18=0.056) + (1/43=0.023)) = 1/0.079 = 12.66Ω.

(top) = 12.66Ω
(mid) = 14Ω
(bot) = 1.5Ω

R(total) = 1/12.66 + 1/14 = 0.079 + 0.071 = 0.150 + 1.5 = 1.65Ω This is where you're going wrong. Think again.

Have I done something wrong? Happy Jack got 8.16Ω.


Resistance Total (Rt) = 1.65Ω
Current Total (It) =
Resistance at A = 6Ω
Current at B =
Power Drawn at C =
Voltage at D =

*Original reply deleted*

Disregard last statement, I have ammended your quote after looking closely at your working out and figured out that it is indeed somewhere else you are going wrong! You're missing a step.

 

[Happy Jack]

Have I done something wrong? Happy Jack got 8.16Ω.

I've updated my original post with some of my workings. It might of some use, there, unless I've made a mistake along the way.

 

[Robson689]

I'll have a go at finishing mine off later tonight, I will have some time to think it all through then

 

[Robson689]

I give up for now

 

[Sintra]

Decided to just start from the top.

Resistance at A = 6Ω (62.6/10.43).

Resistance Total (Rt)

First let's start with the top.

12 + 6 = 18Ω.
1/18 = 0.056Ω.
1/43 = 0.233Ω.

0.056 + 0.233 = 0.289Ω.

1/0.289 = 3.46Ω.

-----

Now the middle.

3.2 + 6.8 + 4 = 14Ω.

-----

Now to tie it all together.

1/3.46 = 0.289Ω.
1/14 = 0.071Ω.
1/1.5 = 0.667Ω.

0.289 + 0.071 + 0.667 = 1.027Ω

1/1.027 = 0.97Ω



Obviously this is getting me nowhere

Why do you add 1.5 ohms at each step of your calculation Happy Jack?

Just look carefully at the positioning of the 1.5 ohm resistor in relations to the others. ie is it connected in parallel to them.

 

[Baker1988]

here are my answers i ait showing my working as it will take ages and it already 3 on a saturday night / sunday mornng i have also not rounded up so it could be accurate but i would normally riund up to 2 decimal places

my answers are

Resistance total (Rt) = 1.224139008 Ohms
Current total (It) = 187.8871586 Amps
Resistance at A = 6.001917546 Ohms
Current at B = 16.42857143 Amps
Power drawn at C = 16.53 KW
Voltage at D = 187.76 Volts

If this is all titally wrong then i blame it on the 6 cans i have had haha

***************EDIT***************

After looking at it again there would be a voltage drop over the 1.5 Ohm resistor so the power drawn at C woud be different my new answer is
Power drawn a C = 11.016 KW

 

[Baker1988]

thinking about it i think i have done this totally wrong ill have another go tomorrow once i get the feed back from the mentors

 

[D Skelton]

here are my answers i ait showing my working as it will take ages and it already 3 on a saturday night / sunday mornng i have also not rounded up so it could be accurate but i would normally riund up to 2 decimal places

my answers are

Resistance total (Rt) = 1.224139008 Ohms Nope
Current total (It) = 187.8871586 Amps Nope
Resistance at A = 6.001917546 Ohms Correct
Current at B = 16.42857143 Amps Nope
Power drawn at C = 16.53 KW Nope
Voltage at D = 187.76 Volts Correct

If this is all titally wrong then i blame it on the 6 cans i have had haha

***************EDIT***************

After looking at it again there would be a voltage drop over the 1.5 Ohm resistor so the power drawn at C woud be different my new answer is
Power drawn a C = 11.016 KW Nope

After looking at it again there would be a voltage drop over the 1.5 Ohm resistor

Which would affect all your previous workings, not just the one remember.

Kudos for giving it a good go mate, you have got the voltage at 'D' right, which confuses me as to how you are struggling with the other answers. When you've got time, stick up your workings out and I'll see if I can help you with the areas you're struggling in, maybe see where you're going wrong

 

[Baker1988]

yeah to be honest i did not work out the D till the very end as i didnt know i needed it till i read it haha its just late and i had a few drinks and it is all just not clicking for me haha im gonna have one more go at it and if it then bed for me and try angain with a fresh head haha

 

[Baker1988]

Hi i have had one more crack at it my answers now are

Resistance total (Rt) = 8.156281838 Ohms
Current total (It) = 28.19912364 Amps
Resistance at A = 6.001917546 Ohms
Current at B = 13.41142857 Amps
Power drawn at C = 2.63 KW
Voltage at D = 187.76 Volts

I could not be bothered to type all my woorking out up so here is a pic of my working out that i did



if it is wrong i will try again tomorrow lol

Im not sure about C and B as they are in series the current is constant so that line will pull current for all the resisistors so it will pull same current through all 3 of them but i dont know if u want the answers for the individual resistor or not

 

[D Skelton]

Ok I can see where you've gone wrong there. Top effort as it's all correct apart from the power drawn at C

From what I can see you have made the mistake of using the voltage at D and multiplied it with the total resistance of the row of three resistors instead of dividing it.

By dividing the voltage (187.77V) by the resistance of those three series linked resistors you get the current (13.41A) through that leg as you already have. You then want to use that current and mulitiply it with the 3.2Ω at C to get the voltage dropped across that resistor (42.92V). It is only now you can use that voltage with the common current present through that leg to determine the power drawn at C. (V x I = P) or (42.92V x 13.41A = 575.55W). Another simple way to find power (P) without a voltage is to use the formula; I[SUP]2[/SUP]R (I x I x R = P) or (13.41[SUP]2[/SUP] x 3.2 = 575.55).

 

[Baker1988]

Ahh I understand I should of know that lol bit of a stupid mistake I know better then that was abit late though haha but the I2R seems a lot easier to work out I should of done it that way lol. O well it was fun giving it a go any way