13.8.1 Potentiometer (Part 1)

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xmphysics

xmphysics

Күн бұрын

Пікірлер: 13
@DerekWong-ri7do
@DerekWong-ri7do 7 ай бұрын
There will be current flowing in E2 only when Vax and Vcd are not equal. E2 will produce current when Vax is too low due to decrease in length, and E2 will absorb current when Vax is too high due to increase in length. This is all due to the fundamental rule that the voltage drops across the paths in parallel should be the same!
@ansonlee3926
@ansonlee3926 Ай бұрын
thanks dude❤❤❤
@debarshisen3364
@debarshisen3364 Жыл бұрын
bro you are god, saved me fr, thankyou
@TinodaisheGanyani-te6ky
@TinodaisheGanyani-te6ky Жыл бұрын
U are talented Sir
@EmmanuelChirongo
@EmmanuelChirongo 9 ай бұрын
Thanks sir I was really looking for this I needed a clear explanation after a practical I carried out now I can attempt my questions well
@xmphysics
@xmphysics 8 ай бұрын
You are most welcome
@a.meforyou
@a.meforyou 9 ай бұрын
Finally understood it
@محمدالسديري-ت7و
@محمدالسديري-ت7و 11 ай бұрын
Thank you so much 👏
@Drivenda1ly
@Drivenda1ly 6 ай бұрын
Perfect
@rgravi
@rgravi 9 ай бұрын
1:13 Hi why is it 1.7V? Shouldnt it be lesser than 1.5 due to ir
@xmphysics
@xmphysics 9 ай бұрын
Firstly, in practice a new 1.5 V battery usually has an actual emf higher than 1.5 V. You can use a voltmeter to verify this. Secondly, the internal resistance is irrelevant at null deflection because no current is flowing through the battery. Which is why the potentiometer always measure the open circuit emf.
@rgravi
@rgravi 9 ай бұрын
hi@@xmphysics thanks but im confused why the current doesnt flow in CD. cuz in a normal parallel circuit, the pd in each branch is equal but current still flows in each branch.
@xmphysics
@xmphysics 9 ай бұрын
@@rgravi Well, in a "normal" circuit, there is only one battery. So the pd in each branch is the same because that one battery pushes out the correct amount of current to make it so. Wheareas here, there are TWO batteries. The driver cell has done the job of setting up a pd across that connected segment of the wire, which happens to match the emf of the secondary cell. So the secondary cell does not have to push out any current.
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