Cheers mate, Your videos got me so interested into physics at GCSE I have chosen to pursue it into A Level this September. Just trying to get a head start into the curriculum over the long break.
Is this part of the syllabus? If not then I think this video is really beneficial as it could be one of those q's that come up in the exam that give you all the info you need but expects you to apply that info to solve s problem
It’s not mentioned explicitly on the specification, but it is exactly the type of thing that could come up as an example in a question. I made it because another teacher asked me to.
At this stage it is still worth it, you have enough time before your exams to make some big improvements. But you need to talk to your teachers so they can give you the best personal advice.
Hi, I think it is because, due to Kirchoff's Second law, all voltages in a closed loop = 0. The 2 resistors on either side and the ammeter make two closed loops respectively, hence, the ratio of voltages would both equal the same. Hope that helped!
I would put it this way. According to Kirchhoff’s second law, the sum of potential difference of the upper route and the sum of potential difference of the lower route is the same (V1+V2=V3+V4=emf). Due to the fact that no current is flowing through the ammeter, V1=V3 and V2=V4. Then we can mess around with the algebra to obtain whatever ratio we like to have, e.g. (V1+V2)/V1=(V3+V4)/V3 to obtain V2/V1=V4/V3.
Basicly,when current not going trow ampemetar,u have same potential on both sides of amperetar,thats why current wont flow,so u can draw same picture without ampemetar, put v2 to be 20 v1 10,for other 2 10 and 5,so when u calculqte eqch side u get 2,u are making statemend that r2/r1=r?/r3 is 2=2,then u calculate potential in places where ampwmetar is conected,and u will get same number,so u will know current is not going trow ampemetar,u picked up good value for R2,then u can cqlculate r? From that formula r2/r1=r?/r3 then r?=(r3×r2)/r1
Cheers mate, Your videos got me so interested into physics at GCSE I have chosen to pursue it into A Level this September. Just trying to get a head start into the curriculum over the long break.
Great stuff, I’ll be adding some information on my website this week to help you prepare for September.
@@PhysicsOnline Ah right I'll be sure to have a check later on cheers
Thanks sir, this has really helped me understand Q26 in the 2019 AQA paper 1
I’ve not actually seen that question paper yet
Is this part of the syllabus? If not then I think this video is really beneficial as it could be one of those q's that come up in the exam that give you all the info you need but expects you to apply that info to solve s problem
It’s not mentioned explicitly on the specification, but it is exactly the type of thing that could come up as an example in a question. I made it because another teacher asked me to.
Thank you hero
A AQA summary of electricity would be very helpful if you could make it😀
I’m filming it tomorrow!
@@PhysicsOnline wonderful😀😀
great explanation! although could you use diodes for R1 and R3 instead of resistors?
How do you derive the 2nd law ratio stuff
You lost me after Kirchoff's first law
Wheatstone bridge is used to measure
(A) high resistance (B) low resistance (C) both (D) potential difference
Please reply
Very goid topic👍
Hi I got a e in as physics 38% and i really thought I done well but do u think its actually worth me doing A2
Pls reply
What do you want to do in the future
At this stage it is still worth it, you have enough time before your exams to make some big improvements. But you need to talk to your teachers so they can give you the best personal advice.
if you wanna put the work in, im in a similar situation so if you wanna talk hmu
did you go through with taking physics in a2? if so, hows it turned out
is this applicable to edexcel exam board
It can be very useful, as questions similar to this sometimes crop up in real exams.
@@PhysicsOnline thank you
Hello! Physics online~ love your video so much. But I was wondering why V2/V1= V?/V3, still don’t know how does it come from🥺
Hi, I think it is because, due to Kirchoff's Second law, all voltages in a closed loop = 0.
The 2 resistors on either side and the ammeter make two closed loops respectively, hence, the ratio of voltages would both equal the same.
Hope that helped!
I would put it this way. According to Kirchhoff’s second law, the sum of potential difference of the upper route and the sum of potential difference of the lower route is the same (V1+V2=V3+V4=emf). Due to the fact that no current is flowing through the ammeter, V1=V3 and V2=V4. Then we can mess around with the algebra to obtain whatever ratio we like to have, e.g. (V1+V2)/V1=(V3+V4)/V3 to obtain V2/V1=V4/V3.
Basicly,when current not going trow ampemetar,u have same potential on both sides of amperetar,thats why current wont flow,so u can draw same picture without ampemetar, put v2 to be 20 v1 10,for other 2 10 and 5,so when u calculqte eqch side u get 2,u are making statemend that r2/r1=r?/r3 is 2=2,then u calculate potential in places where ampwmetar is conected,and u will get same number,so u will know current is not going trow ampemetar,u picked up good value for R2,then u can cqlculate r? From that formula r2/r1=r?/r3 then r?=(r3×r2)/r1
🎶✈️✈️