Physics - Electricity / Circuits
Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms? 
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
How have you worked these out? I don't think so. What you want to do in both cases is:
1) Find all paths between the two points (there will be two routes between each pair, one going clockwise and the other counterclockwise)
2) Find the resistance for each path
3) Work out the total resistance
Original post by ohnanailikenanas
Sorry I'm posting quite a few questions but I'm only trying to learn and see if I'm doing this right! Would anyone be able to tell me if this is right? And would (a) (ii) X and Z be 8 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
Two rules:
1) Series resistors are summed: R1 + R2 + R3 etc.
2) Parallel resistors are:
In part a) think of the paths current will take flowing between the two points in question.
In the first there are two paths which means these paths are in parallel:
1) The first is directly between X and Y flowing through the 6 ohms resistor on its own.
2) The second flows via 3 resistors passing through each in turn (series) 6, 3 and 12 ohms.
To calculate the resistance between the terminals, calculate the resistance between each path independently (the series paths). Replace each of the series paths with a single equivalent resistance.
You then have only two resistances to deal with and both are in parallel.
The final resistance can now be calculated.
(edited 9 years ago)
Hm, I thought that's what I was doing 
Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/
Posted from TSR Mobile
Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/ Posted from TSR Mobile
Original post by ohnanailikenanas
Hm, I thought that's what I was doing Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/
Posted from TSR Mobile
Since I have to find the resistance b/w X and Y.. It's in parallel with the 3 Ohms so do I not do 1/3 + 1/6 = 4/12 + 2/12 = 6/12 then flip 12/6 = 2 Ohms??? And then between X and Z is it not 6 Ohms because the 6 Ohms is parallel to the 12 Ohms which equals to be 4 Ohms then if the other is 2 Ohms add it together to make 6 Ohms? :/ Posted from TSR Mobile
It's not in parallel with the 3Ohms only though- it's in parallel with the entire anti-clockwise route(6,3,12). You can find the resistance of that by combining them using the rules for series circuits.
Wait.. Are you trying to explain to me that b/w X and Y you count it as if 6, 3, 12 Ohms then the shorter way 6 Ohms? Do I still use the parallel equation thing? Eg 1/6 + 1/12 etc
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
Wait.. Are you trying to explain to me that b/w X and Y you count it as if 6, 3, 12 Ohms then the shorter way 6 Ohms? Do I still use the parallel equation thing? Eg 1/6 + 1/12 etc
Posted from TSR Mobile
Posted from TSR Mobile
yes. the longer path is 6,3 and 12 ohms. These resistors are in series.
The equivalent resistance is 6 + 3 + 12 = 21 ohms.
This equivalent is in parallel with the 6 ohms resistor directly connected between X and Y.
21 ohms equivalent is in parallel with 6 ohms.
So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
You should round your answer to 4.67 ohms or state it exactly as
(edited 9 years ago)
Oh okay, great! Thank you so much! I'll try and work out the second question now.
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
So the resistance b/w X and Y is 1/21 + 1/6 = 2/42 + 7/42 = 9/42 = 42/9 = 4.66 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
Now you know how to do it, what did you make the answer between X and Z?
I did 6 + 3 = 9 then 6 + 12 = 18. 1/9 + 1/18 = 4/36 + 2/36 = 6/36 = 36/6 = 6 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
I did 6 + 3 = 9 then 6 + 12 = 18. 1/9 + 1/18 = 4/36 + 2/36 = 6/36 = 36/6 = 6 Ohms?
Posted from TSR Mobile
Posted from TSR Mobile
Ah thank you ever so much! You don't realise how much I appreciate this!
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
Sorry, again. Is part (b) right? 
Posted from TSR Mobile
Posted from TSR Mobile
Original post by ohnanailikenanas
As before, there are two current paths between X and Z and we must consider the path with the 3 ohms resistor in it.
I3ohms = VXZ / RXZ
That is because all of the series current flowing in that path must flow through the 3 ohms resistor.
So
VXZ = 12V
RXZ = total resistance between XZ (current path with 3 ohms resistor in it) = (3 + 6)ohms = 9 ohms
I3ohms = 12 / 9 = 1.33 amps.
(edited 9 years ago)
Ah okay I see, thank you once again!
Posted from TSR Mobile
Posted from TSR Mobile
Quick Reply
Related discussions
- Use a simulator to plot the frequency response of the circuit
- physics alevel edexcel circuit question
- BTEC Applied science Unit 3 2022 Exam
- Help! Electricity igcse physics qs make no sense
- LDR's in GCSE physics
- need help please!
- Things to do in gap years?
- Physics Alevel question :- electromagnetism induction and alternating current
- Question about parallel circuit- OCR AS physics
- electric circuit question
- resistance in a parallel circuit
- A level physics electricity
- URGENT: What equations ARE NOT on the AQA GCSE Physics equation sheet?
- Physics Aqa GCSE help
- I don't know what subject to pick
- Edexcel GCSE Physics Paper 2 Higher Tier Triple 1PH0 2H - 16th June 2023 [Exam Chat]
- How does the variable resistor circuit provide a larger current?
- Impedance
- Electrical
- Studying an Engineering Access to HE course after gap year.
