How does Terminal PD get reduced when greater I is withdrawn? V=IR, so if I is greater, the V should also be greater. It doesn't make sense to me. Please explain me this one
Well i understood it..... It's because of the internal resistance that gets increased when R withdraws more current so the terminal PD gets lower as a result.
ε = V + Ir emf is a measure of the maximum p.d. a battery can supply, so it must stay the same since you're always using the same battery. If you increase I, then Ir obviously increases. V must therefore decrease if you want emf to be the same. Essentially ε = V + Ir so if Ir increases then V must decrease in order for ε to stay the same.
I think its because in order to Increase the Current you are reducing the resistance of the variable resistor and therefore the total resistance of the circuit which is given by R. So I guess these changes cancel out to give the same value
+A Level Physics Online Aha. Your videos are incredibly useful :) I'm starting to enjoy this topic a lot more than I ever did at GCSE, the way you go through the content in your videos is also interesting so I thank you for taking the time to do that.
that is great to review the basics of elec course
those battery cells look delicious just saying, but anyway very helpful vid ty!
They do be looking like some sort of biscuit with chocolate lol
Reminds me of cat food (bad) not gonna lie.
these videos have really helped me with my exam tomorrow- thankyou!
How did you do? Do you have your results yet?
nah she still in the exam hall
i really like your videos .. its easy to understand A level physics lectures over here .. :)
Thank you!! Saved my mocks when i missed the lesson on internal resistance
Please keep posting more videos. Thankyou!
THANK YOU!
Awesome
How does Terminal PD get reduced when greater I is withdrawn? V=IR, so if I is greater, the V should also be greater. It doesn't make sense to me. Please explain me this one
Well i understood it..... It's because of the internal resistance that gets increased when R withdraws more current so the terminal PD gets lower as a result.
ε = V + Ir
emf is a measure of the maximum p.d. a battery can supply, so it must stay the same since you're always using the same battery.
If you increase I, then Ir obviously increases. V must therefore decrease if you want emf to be the same.
Essentially ε = V + Ir so if Ir increases then V must decrease in order for ε to stay the same.
Nice one
@@mattwhitelock4725 couldn’t you use that same logic and say if I increases, IR also increases? then the Ir is the one that decreases
I think its because in order to Increase the Current you are reducing the resistance of the variable resistor and therefore the total resistance of the circuit which is given by R. So I guess these changes cancel out to give the same value
I love your drawing
Could you use a multimeter to find the internal resistance of the battery using the resistance setting?
yeah i have this same questison
why is V/I not equal to R(external resistance) since V is actually the p.d across the resistor?
Thankyou
One thing I don't understand is the terminal p.d. Is the terminal p.d the same across the variable resistor?
Eloise CP i would guess so since terminal p.d is the external voltage of a circuit.
Thanks a lot!! :D
+Michael Harding No problem? It was only recently I actually had a look inside a large battery to see what it was actually made from.
+A Level Physics Online Aha. Your videos are incredibly useful :) I'm starting to enjoy this topic a lot more than I ever did at GCSE, the way you go through the content in your videos is also interesting so I thank you for taking the time to do that.
Thanks
jaffa cakes
Burnnnn vocabulary