Yes there are 2 parallel routes taking you from X to Y. One with a resistance of 60 (30+30) and the other with a resistance of 40 (20 + 20), hence the use of the parallel formula
for qn 3 i calculated 1/120+1/500 and i didnt got 96.77 i got a different answer, i got 31/300 which is 0.0103 in decimals. what do i do wrong how did you got 96.77 ohms.
Hi. Whilst you've stated in the comments that these are AQA past questions I do not recall ever seeing questions involving mean drift velocity (as seen in Q12). Can you clarify this please? Many thanks
If you see the publish date, it was from 2018. Highly possible that AQA have now removed this from their specification. I haven't taught AQA for a few years so I wouldn't know if they have or not! It is a useful concept to know about even if it isn't in the specification though
It would change, making the resistance fixed allows us to revive a single answer if it wasn’t, we wouldn’t be able to get an answer as there is no value of resistance we can use
Potentially but typically an exam comes after having explained concepts in class and I felt going into more detail would have simply made the video too long!
But they aren't being multiplied or divided here. When we have resistors in parallel we add their reciprocals and then reciprocal the answer. The approach here used the fact that percentage uncertainty means that the value could be 10% higher or 10% lower. I simply figured out how much bigger than the calculated value the total resistance COULD be and found it to be about 9% different
For 6 wouldn't it be A because the three resistors gives the highest resistance, and the question is asking which one dissipates (i.e wastes) the most?
High resistance doesn't mean you waste the most energy. You have to calculate the power dissipated by a resistor to do that which is proportional to current squared
COULD U HELP ME WITH THIS QUESTION PLEASE a cabel consists of 10 strands of copper wire each of cross-section area 1.1x10^-3 cm^2. Calculate (a) the resistance per meter of the cable (b) the minimum number of strands which would be required if the resistance per meter is not to exceed 0.010ohms. (resistivity of copper = 1.8 x 10^-8 ohm meter.)
Multiplying by 1.1 calculates the resistance if each of the resistors were 10% larger than the stated value. Using those 10% higher values calculate the largest possible combined resistance hence why Rmax is used
Essentially you have 2 parallel branches connecting X to Y. (One branch with a total resistance of 30 + 30 = 60 and the other branch with a total resistance of 20 + 20 = 40) Since those two branches are in parallel, we calculate their total resistance using: 1/R = 1/40 + 1/60
This was genuinely so helpful, thankyou so much because I can never seem to find good exam questions on youtube.
Sir I love your videos!!! Thank you so much for the concise and organised explanations/model answers!!!
rock on dude
1:41, Why did you use the parallel formula for the two 30 ohms resistors? Are they in parallel the two 30 ohm resistors?
Yes there are 2 parallel routes taking you from X to Y. One with a resistance of 60 (30+30) and the other with a resistance of 40 (20 + 20), hence the use of the parallel formula
11:45 in the final step shouldn’t it be 3x 230^2??
Yes I believe you are correct! Thanks for flagging!
You're making the calculations very complicated and this is gonna confuse the students. I don't recommend this video
Which video do you reccomend?
As a student, I think the calculations are quite clear to understand and found this video very useful. I will be recommending my friends to watch it
In the question at 17:42 can not solve it with the E.M.F = V+Ir formula, because when I try to solve it the r comes 0 ohms?
Hi is dis for ocr a?
pretty sure this is for AQA A but the electricity questions are pretty much the same in both
Thank you so much !!
for qn 3 i calculated 1/120+1/500 and i didnt got 96.77 i got a different answer, i got 31/300 which is 0.0103 in decimals. what do i do wrong how did you got 96.77 ohms.
Did you remember to take the reciprocal once you'd added them together? Remember 1/R total = 1 / R1 + 1 / R2... + ....
Hi. Whilst you've stated in the comments that these are AQA past questions I do not recall ever seeing questions involving mean drift velocity (as seen in Q12). Can you clarify this please?
Many thanks
If you see the publish date, it was from 2018. Highly possible that AQA have now removed this from their specification. I haven't taught AQA for a few years so I wouldn't know if they have or not! It is a useful concept to know about even if it isn't in the specification though
On question 11 shouldn’t it be R= 3*230squared/10500 ?
Yes it should be but his answer is still right so he just put in the wrong number
Yes, you are right
question 11 is wrong, you're meant to do 3*230^2/10500 not 2*230^2/10500
His answer is right he just put a 2 instead of a 3
in 15:09 what would happen to the reading on the voltmeter if the resistance in the variable resistor was not fixed??
It would change, making the resistance fixed allows us to revive a single answer if it wasn’t, we wouldn’t be able to get an answer as there is no value of resistance we can use
good questions but you it woulve been better if you explained some of the concepts along the way too
Potentially but typically an exam comes after having explained concepts in class and I felt going into more detail would have simply made the video too long!
Why do you not add the percentage uncertainties together for question 3, that's what we've been told when multiplying and divide numbers,
But they aren't being multiplied or divided here. When we have resistors in parallel we add their reciprocals and then reciprocal the answer.
The approach here used the fact that percentage uncertainty means that the value could be 10% higher or 10% lower. I simply figured out how much bigger than the calculated value the total resistance COULD be and found it to be about 9% different
@@burrowsphysics thank you very much
@@burrowsphysics I had the idea that since we’re having to take the reciprocal we’re basically dividing but that’s not really the case
For 6 wouldn't it be A because the three resistors gives the highest resistance, and the question is asking which one dissipates (i.e wastes) the most?
High resistance doesn't mean you waste the most energy. You have to calculate the power dissipated by a resistor to do that which is proportional to current squared
I don’t understand how you got 0.8 for the percentage question 9 .
0.4%+0.4% because you are doing 0.5mx0.5m
COULD U HELP ME WITH THIS QUESTION PLEASE
a cabel consists of 10 strands of copper wire each of cross-section area 1.1x10^-3 cm^2. Calculate (a) the resistance per meter of the cable (b) the minimum number of strands which would be required if the resistance per meter is not to exceed 0.010ohms. (resistivity of copper = 1.8 x 10^-8 ohm meter.)
@Yung PouriiP rude
isnt the anwser to question 8 D and not B
No you need the thermistor to have very high resistance and the LDR to have low resistance to maximise the voltmeter reading which is why D is wrong
why did you multiplied 1/120 and 1/500 with 1.1. And what does 1/R max means
Multiplying by 1.1 calculates the resistance if each of the resistors were 10% larger than the stated value. Using those 10% higher values calculate the largest possible combined resistance hence why Rmax is used
im lost on question 8 tbh
loool truss where does ldr's come in
Can someone explain the equation used in question 8?
That's just the potential divider equation that you can use to calculate the potential difference across one of the resistors in series
7:50
in Question 1 you don't have to find the total power in parallel like we do with resistance 1/R + 1/R ???????
I hate uncertainty question 😕
I don't think it is anyone's favourite part of being a scientist but it is an important part of all experimental work!
Can someone explain question 2 please?
Essentially you have 2 parallel branches connecting X to Y. (One branch with a total resistance of 30 + 30 = 60 and the other branch with a total resistance of 20 + 20 = 40)
Since those two branches are in parallel, we calculate their total resistance using: 1/R = 1/40 + 1/60
Because you can't add 2 fractions with a different denominator, we have to find the lowest common in so we x one by 2 and the other by 3