GCSE level Electricity covering: Potential Dividers and how to divide the input voltage to produce an output voltage which is lower than the input voltage.
@@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 ?
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.
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?
If the voltage is 0v on the other side of resistance2 .how does current still flow until it reaches the battery? There is no voltage to push the charge. By the way your videos are really helpfull thank you
DrPhysicsA just across the second resistor, the voltage drops to 0v. So how does current continue on flowing to reach the battery if there is no voltage between R2 and the negative side of the battery. The vltge at the negative terminal is 0v and the vltge just across R2 is also 0v. There is no potential difference
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.
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.
This helped so much my textbook/teacher was terrible at explaining this!
Your videos are AMAZING!! i am doing IGCSE and they are still reaaallly helpful. Thank you :)
Thank You very much !
It was very clear and well explained...
From Sri Lanka...
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
Really helpful, thank you.
Sir, you are a genius! Thank you for such a nice explanation, I finnaly understood it!
life saver exam saver future saver physics saver everything saver!
AMAZING, I FINALLY UNDERSTAND! best teacher
excellent - thanks
Very well explained. Helped me so much. Thank you :)
What are your plans next, are going to make new videos?
thank you so much!!!!
Thanks a lot
U made my day
Thank you 😊
best teacher and physics channel shame u dont make anhmore hope.u dp
very well explained!!! thanks
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 ?
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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...
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).
So helpful 👌🏽👌🏽
Thanks you saved me
Thank you so much
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.
Legend
If the voltage is 0v on the other side of resistance2 .how does current still flow until it reaches the battery? There is no voltage to push the charge. By the way your videos are really helpfull thank you
Thanks. Whereabouts does this happen?
DrPhysicsA just across the second resistor, the voltage drops to 0v. So how does current continue on flowing to reach the battery if there is no voltage between R2 and the negative side of the battery. The vltge at the negative terminal is 0v and the vltge just across R2 is also 0v. There is no potential difference
DrPhysicsA during your example with 12v battery and 2 6ohm resistors
Adam Asaad lol he never answered
all i can think about is star wars amongst so many mentions of R2...
very helpful
Thanks.
THANK U
soooo useful
it sounds good now get practical the load resistance can not be less than the r1 value or there will be a voltage drop.
If it was the other way around where the filament lamp is connected to the fixed resistor not to the input transducer
Why there are 4 dislikes? Lol
i'll never understand this shit and my ig's exam is in 2 hours