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I think it was 0.2 A, but I don't remember for sure..
Emf was 2 V.
How was the emf 2V ? How did you get that answer ? I got 1.8V and 2.1* 10^-3 for fringe separation.
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I think it was 0.2 A, but I don't remember for sure..
Emf was 2 V.
Well since galvanometer reading is 0 so emf in cell B has to be equal to emf in cell A right? :SHow was the emf 2V ? How did you get that answer ? I got 1.8V and 2.1* 10^-3 for fringe separation.
Well since galvanometer reading is 0 so emf in cell B has to be equal to emf in cell A right? :S
and that only happens when emf in one cell is equal to the one in another. :/No that just shows that no current is flowing through the galvanometer
.. That would kinda defeat the point of the potential divider though if you were using two cells of equal emf.. then there would be no need to shift the pointer to get a galvanometer reading of 0.. it would already be 0 at the maximum distance.. you're trying to find the emf.. you get to the point where no current is flowing so that you can ignore the internal resistance of the cell and find an accurate value for emf..and that only happens when emf in one cell is equal to the one in another. :/
Dude u just rock!!!!i got the same.was the e.m.f of B 1.03V?
actually the main thing tht works behind this galvanometer giving zero is the kirchoff's law itself... That would kinda defeat the point of the potential divider though if you were using two cells of equal emf.. then there would be no need to shift the pointer to get a galvanometer reading of 0.. it would already be 0 at the maximum distance.. you're trying to find the emf.. you get to the point where no current is flowing so that you can ignore the internal resistance of the cell and find an accurate value for emf..
Yea I know.. that is the principle I applied.. but I got 1.8V as the Pd 0.9/1 x 2v but you might be right because my method does not account for the other resistorsactually the main thing tht works behind this galvanometer giving zero is the kirchoff's law itself.
when u consider the loop below, the potential difference on the wire above has a positive value while the emf (at B in this case) has a negative value, so that when u sum it up u get a resultant potential difference.
so to make the galvanometer give a zero reading, the potential difference on the wire MUST be equal to the emf( at B in this case).
thus if u are able to find the pd across the respective length of wire which gave a zero reading, then u actually found the emf of B!!!!
but if it was to 2V like cell A the balance point would have been at the end of the junction i had it 1.8 V....anyway i dun thimk 4 marks matter that muchI think it was 0.2 A, but I don't remember for sure..
Emf was 2 V.
thats what i did ALHAMDULILAH,Guys I'm sure the emf was 1.03V even though I put 1.8V.. you couldn't have used the equation of the ratio of lengths multiplied by emf of battery because it assumed there are no other sources of resistance besides the wire.
You would I guess use the R(across length of 0.9m)/(total resistance) and you'd get 1.03V ... I think it was meant to be tricky like that..
So im guessing we'd get no marks at all for using the equation with the ratios either because we missed the point of the question so 3 marks lol.. no biggie
was it elastic or inelastic?thats what i did ALHAMDULILAH,
obviously inelastic as ke not conserved even you can write speed of seperation is not equal to speed of approach i remebr one was less and one was more so i wrote speed of aproach or seperation dont remeber but which one of these was more was more than speed of seperation or approach dont remebr which one of these but which was less.was it elastic or inelastic?
soory i dont remeber but i suppose theres no need to discuss now most of the people here would have forgot the answes.can anyone plz tell me what was the first part of circuits question in the paper..
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