@@nychan2939 DrPhysicsA is brilliant & by comparison, most people suck at explaining things. 1 of my teachers taught the class that electrons are neutral. I had to correct him in front of the class. I ended up getting expelled.
Thank you DrPhysicsA for teaching gcse physic, as a self-study student myself I find it very useful to learn from your video. I'll look forward for more of your teaching video!
Thank you sir for your efforts. Won't the 6 volts lamp attached in 11:47 decrease the overall resistance of the resistor to which it is attached in parallel allowing a comparatively higher value of current in the circuit ?
I have a question, in the exam if you work out the Resistor values and you put 2/3 , then I say that R1 is 2ohm and R2 is 1ohm then would that be incorrect? because how would you know whether it's 20/30 ohm or 2/3 ohms? Hope I'm being understanding to my question.
+Sean o'connell Well you are right that if all you know is the ratio of the resistor values you cannot determine the actual value of each resistor. But you would usually be given some other info (such as current).
Great video as always, thanks! I have a query. My textbook says that potential dividers cannot be used to light bulbs. It shows an example in the book. If you have a 6V supply and 2 resistors (R1 and R2) each of which is 100 ohms and placed in series then the p.d. drop across each is 3V. It says a typical 3V bulb has a resistance of 15 ohms. If we put this across R1 for example, the equivalent resistance will be 13.04 ohms. Then it calculates the new voltage across R1: V1 = (R1/Rtotal) Vtotal......V1 = (13.04/113.04) 6......V1 = 0.7V. So it says that the bulb doesn't light because as soon as you connect it across R1 this changes the mechanics of the entire circuit and the bulb only gets 0.7V. Is this right? If so, what is the point of a potential divider if as soon as you put a load across the resistor you no longer get that voltage?
You can't put a filament bulb across R2 as you have changed the resistance in the divider. Well explained (and beautiful videos by the way). But perhaps R2 should have been the filament bulb itself as a better example.
Thank you very much for making so good videos, I'm having tomorrow an exam and I totally understood it!!!!
-Diana from Spain
+dianeixion Thanks. Hope the exam goes really well.
my physics teacher doesn't teach our class at all... this really helped!!
Rhea K zz
Why so many people blame their teachers?
@@nychan2939 DrPhysicsA is brilliant & by comparison, most people suck at explaining things. 1 of my teachers taught the class that electrons are neutral. I had to correct him in front of the class. I ended up getting expelled.
Your videos are AMAZING!! i am doing IGCSE and they are still reaaallly helpful. Thank you :)
This helped so much my textbook/teacher was terrible at explaining this!
Thank you DrPhysicsA for teaching gcse physic, as a self-study student myself I find it very useful to learn from your video. I'll look forward for more of your teaching video!
This (+ your channel) has literally made my day/week/Gcse Physics
Thank You very much !
It was very clear and well explained...
From Sri Lanka...
Thank you sir for your efforts. Won't the 6 volts lamp attached in 11:47 decrease the overall resistance of the resistor to which it is attached in parallel allowing a comparatively higher value of current in the circuit ?
Sir, you are a genius! Thank you for such a nice explanation, I finnaly understood it!
AMAZING, I FINALLY UNDERSTAND! best teacher
What are your plans next, are going to make new videos?
life saver exam saver future saver physics saver everything saver!
I have a question, in the exam if you work out the Resistor values and you put 2/3 , then I say that R1 is 2ohm and R2 is 1ohm then would that be incorrect? because how would you know whether it's 20/30 ohm or 2/3 ohms? Hope I'm being understanding to my question.
+Sean o'connell Well you are right that if all you know is the ratio of the resistor values you cannot determine the actual value of each resistor. But you would usually be given some other info (such as current).
Very well explained. Helped me so much. Thank you :)
Really helpful, thank you.
Great video as always, thanks! I have a query. My textbook says that potential dividers cannot be used to light bulbs. It shows an example in the book. If you have a 6V supply and 2 resistors (R1 and R2) each of which is 100 ohms and placed in series then the p.d. drop across each is 3V. It says a typical 3V bulb has a resistance of 15 ohms. If we put this across R1 for example, the equivalent resistance will be 13.04 ohms. Then it calculates the new voltage across R1:
V1 = (R1/Rtotal) Vtotal......V1 = (13.04/113.04) 6......V1 = 0.7V.
So it says that the bulb doesn't light because as soon as you connect it across R1 this changes the mechanics of the entire circuit and the bulb only gets 0.7V. Is this right? If so, what is the point of a potential divider if as soon as you put a load across the resistor you no longer get that voltage?
Quite right. It means you have to choose the values of R1 and R2 carefully.
best teacher and physics channel shame u dont make anhmore hope.u dp
Thanks a lot
U made my day
You can't put a filament bulb across R2 as you have changed the resistance in the divider. Well explained (and beautiful videos by the way). But perhaps R2 should have been the filament bulb itself as a better example.
That's what I thought...
very well explained!!! thanks
excellent - thanks
thank you so much!!!!
So helpful 👌🏽👌🏽
Thank you 😊
If it was the other way around where the filament lamp is connected to the fixed resistor not to the input transducer
Thanks you saved me
THANK YOUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU!
it sounds good now get practical the load resistance can not be less than the r1 value or there will be a voltage drop.
Thank you so much
all i can think about is star wars amongst so many mentions of R2...
Legend
very helpful
soooo useful
THANK U
Thanks.
Why there are 4 dislikes? Lol
i'll never understand this shit and my ig's exam is in 2 hours