You are the best! I am self studying since online classes don't work for me(Due to the strict routine and all that work load doesn't get me motivated which used to during physical classes). But with your help, I am progressing pretty well. Thank you!
This video was excellent! Such a simple and straightforward explanation! I was struggling with this concept when I first learned it but after this video I am no longer confused!
I don’t disagree but I find using ratios a much more efficient way I mean this way works aswell but some questions it can be really tricky to use this formula where using ratios is much easier. Still a great video and your making a level a lot easier then it is👍
Is the output voltage always the same, regardless of what you have in the circuit? Like is it always just the emf for a supply with negligible internal resistance? Or does it increase if you have more resistance in the circuit? Is it just current that varies?
I don't understand how when R2 is increased,V(out) also increases but in the equation R2 in both the numerator and denominator increases by the same amount so why does V(out) increase and not stay equal
Why only R2? Why didn't you consider R1? Why is R2 so special in this case that only it has a voltmeter around it? Edit: What I mean to say is, shouldn't there also be I=Vmid / R1 ?
Now this makes sense to me.. for some reason I've been told that a higher resistance will give a smaller share of the terminal p.d. and just makes no sense... AT ALL...
I think its because maybe they were describing it if R1 had a high resistance, R2 would have a smaller voltage drop as a result but that's kinda convoluted imo.
Of course it does. Because every cell/battery will have some amount of internal resistance. This equation wouldn't be very helpful if it only applied to theoretical circuits.
Nah - V in is being given to the resistors and is being shared between the two. V out is the portion that Resistor 2 gets. You could connect a load, such as a light, to Resistor 1 as well, then you've got 2 V outs (one on R1 one on R2). If you did this and added the voltage value for both V outs, you would get your V in [Kirchov's second law] :) Its easier to think aboot with a diagram
I've been looking everywhere but I still can't find an answer to my question. I am currently studying the OCR AS level (spec A) physics course and there's a part on "Wheatstone circuits with a strain gauge": In the book they give you an equation for calculating the output voltage of the circuit, but it has missing symbols :( (there's an error in the textbook). I asked my teacher but he doesn't know lol, could you please help me? gyazo.com/631086e3570a4c55bb9ee57d8c0067b3 (there's the picture of the equation and circuit)
+awesomeoverawesome Ok, I'll try and answer this in words - it may be best to write it out yourself too. Question 2. Resistance = (Rho x Length) / Area Volume = Area x Length so therefore Area = Volume / Length Resistance = (Rho x Length) / (Volume / Length) Resistance = (Rho x Length Squared) / Volume This means if the resistivity and volume remain constant then Resistance is proportional to Length squared
+awesomeoverawesome Answer to question 3. Initially when all resistors are 100 ohms the circuit is 'balanced' and the output voltage is 0. Because V = ((100/200) - (100/200))x 6 = 0 When there is a 0.001% change to 100.1ohms across R1 then the pd is equal to 0.0015V Is that right? en.wikipedia.org/wiki/Wheatstone_bridge
A Level Physics Online Thank you so much for all your responses, I can't put into words how much I appreciate it! Typical of the OCR textbook, there are no answers to this part :( This section is found in the summary part, but it is still part of the spec, and there are no answer online either. Looking at question 3, isn't a 0.001% increase/decrease of 100 = +- 0.001? So let's say the resistor was to increase by 0.001%, R1 would have a resistance equal to 100.001 ohms. So applying the equation: V = ((R1/(R1+R2)) - ((R4/(R3+R4)) x 6.0 V = ((100.001/(100.001+100)) - ((100/(100+100)) x 6.0 V = 1.5 x 10^-5 V Is that right? Or did I go wrong?
+awesomeoverawesome But don't worry too much about this question - the Wheatstone bridge is not in the OCR spec - it is only an extension task in the book.
That was so simple, you're really positive aswell, I feel like everything's going to be okay
My physics teacher is pretty shite. So you're the real MVP. Cheers bruv
my teacher too is shit. You are better.
ur eng teacher is shit* too, get it?
I also have a really good physics teacher and a really bad one, and I have the bad one for electricity
@@grassytramtracks i have bad teachers for both parts of physics💀
In class, I was trying to derive it, couldn't get anywhere. I come on TH-cam and it's explained by this awesome dude in less than 5 mins
You are the best! I am self studying since online classes don't work for me(Due to the strict routine and all that work load doesn't get me motivated which used to during physical classes). But with your help, I am progressing pretty well. Thank you!
Even in 2020 I’m watching these videos and they are so great - thank you so much
Oh my, watching this the morning of my exam and NO lesson ive ever had has explained it this well, thank you!!!
Wow that was so clear
Thank you very much, couldn't understand potential dividers at all until now.
Thank-you, you've really been a big help through this lockdown and have been my saviour so many times.
I have an electricity test tomorrow and you have fixed my one weakness in 4 minutes - thanks !!
Ty, this was very helpful. There are a bunch of questions on this in IGCSE physics yet not a single textbook covers this topic
i luv this!! he showed how the formula was derivedd!!!!!! thank you!!!
This video was excellent! Such a simple and straightforward explanation! I was struggling with this concept when I first learned it but after this video I am no longer confused!
Thanks mr matheson, doing this in my AS mocks and this video helped so much - big dad lucas (KS)
Not a problem
thank you so much you just saved me like 5-6 marks
I don’t disagree but I find using ratios a much more efficient way I mean this way works aswell but some questions it can be really tricky to use this formula where using ratios is much easier. Still a great video and your making a level a lot easier then it is👍
thanks thats epic dude, bro, 99
for some reason turning this on its side like in this video has made it a million times clearer
this made it so simple omg
Excellent explanation!
Thank you for your explanation, although I am a Dse candidate but not a A-Level candidate, this video is still helpful to me🙏
What is dse
@@jesselingardinho1374 A public exam in Hong Kong
@@II-ig5nt do u guys have the same syllabus coz I heard hk students talking abt it in the uk
@@jesselingardinho1374 i think the syllabus of Physics in HK is not as wide as A-Level
im not in Alevels but idk for some reason, his videos give me an aesthetic vibe. like the pens he uses and the writing.
Kuddos to you big sir, you're the best I've seen some far :) keep up the good work. Cheers
cleared things up so much, thanks
got the paper 1 a level tomorrow 9am it is currently 11pm and i'm going to be doing an all nighter but it fine because I got my red bulls
Thanks very much!
thank you so much this was really helpful❣❣
Is the output voltage always the same, regardless of what you have in the circuit? Like is it always just the emf for a supply with negligible internal resistance? Or does it increase if you have more resistance in the circuit? Is it just current that varies?
thanks
THANK YOU SIR
You can also make videos for Olevels students 🙂
I don't understand how when R2 is increased,V(out) also increases but in the equation R2 in both the numerator and denominator increases by the same amount so why does V(out) increase and not stay equal
If R2 changes Vout and V1 both should change, but R1 is fixed, does that mean I changes?
Thank you!!!
THANKS A LOT SIR.
Your awesome.
Why only R2? Why didn't you consider R1? Why is R2 so special in this case that only it has a voltmeter around it?
Edit: What I mean to say is, shouldn't there also be I=Vmid / R1 ?
You are the GOAT 🐐
If only you posted this before my Unit 1 exam :(
aleryani 98
He posted this 2 years ago.. you commented 1 year ago... BOI.
What happens if an additional resistor is added
V in is the emf right?
Yes
is there any video about potentiometer
Saimakhan Khan
I'm doing the exam tomorrow! :0
Same :( Good luck!
Now this makes sense to me.. for some reason I've been told that a higher resistance will give a smaller share of the terminal p.d. and just makes no sense... AT ALL...
I think its because maybe they were describing it if R1 had a high resistance, R2 would have a smaller voltage drop as a result but that's kinda convoluted imo.
what was that at 1:49 to 1:52 ??
+Dr. Fhood Kirchhoff's First Law: th-cam.com/video/pJ7xrmUb_UQ/w-d-xo.html
A Level Physics Online thanks
kirchoff's first law
does this still apply if the cell has internal resistance
Of course it does. Because every cell/battery will have some amount of internal resistance. This equation wouldn't be very helpful if it only applied to theoretical circuits.
yes. The resistors are still getting some voltage, even though it's less than emf. In our scenario here, " V in" is terminal voltage of cell :)
My AS level physics is in less than 2 hours 😭
You’ll be fine!
What happens to the current if the resistance changes
everything fucking blows up
GCSE GANG
IGCSE GANG 💪💪😍😍
Will Vout be more than Vin?
Nah - V in is being given to the resistors and is being shared between the two. V out is the portion that Resistor 2 gets.
You could connect a load, such as a light, to Resistor 1 as well, then you've got 2 V outs (one on R1 one on R2). If you did this and added the voltage value for both V outs, you would get your V in [Kirchov's second law] :)
Its easier to think aboot with a diagram
I've been looking everywhere but I still can't find an answer to my question.
I am currently studying the OCR AS level (spec A) physics course and there's a part on "Wheatstone circuits with a strain gauge":
In the book they give you an equation for calculating the output voltage of the circuit, but it has missing symbols :( (there's an error in the textbook).
I asked my teacher but he doesn't know lol, could you please help me?
gyazo.com/631086e3570a4c55bb9ee57d8c0067b3
(there's the picture of the equation and circuit)
+awesomeoverawesome Ok, I'll try and answer this in words - it may be best to write it out yourself too.
Question 2.
Resistance = (Rho x Length) / Area
Volume = Area x Length so therefore Area = Volume / Length
Resistance = (Rho x Length) / (Volume / Length)
Resistance = (Rho x Length Squared) / Volume
This means if the resistivity and volume remain constant then Resistance is proportional to Length squared
+awesomeoverawesome Answer to question 3.
Initially when all resistors are 100 ohms the circuit is 'balanced' and the output voltage is 0. Because V = ((100/200) - (100/200))x 6 = 0
When there is a 0.001% change to 100.1ohms across R1 then the pd is equal to 0.0015V
Is that right?
en.wikipedia.org/wiki/Wheatstone_bridge
A Level Physics Online Thank you so much for all your responses, I can't put into words how much I appreciate it!
Typical of the OCR textbook, there are no answers to this part :( This section is found in the summary part, but it is still part of the spec, and there are no answer online either.
Looking at question 3, isn't a 0.001% increase/decrease of 100 = +- 0.001?
So let's say the resistor was to increase by 0.001%, R1 would have a resistance equal to 100.001 ohms.
So applying the equation:
V = ((R1/(R1+R2)) - ((R4/(R3+R4)) x 6.0
V = ((100.001/(100.001+100)) - ((100/(100+100)) x 6.0
V = 1.5 x 10^-5 V
Is that right? Or did I go wrong?
+awesomeoverawesome You're correct. I didn't read the question properly!
+awesomeoverawesome But don't worry too much about this question - the Wheatstone bridge is not in the OCR spec - it is only an extension task in the book.
i understood it until i attempted the eq my teacher set lmfao
3:39 "If you know the current and the resistance at any point in the circuit, you can work out the potential difference across that b*tch"
the video is great youre amazing etc etc etc.... but oh my god your intro sound. why. what. stop.
I have stopped, it’s not there in my videos from the last couple of years
What a guy ❤️
Lol...its in o level for me :v