1correct ans is 8.33 volts.........
i need solution
the cells given in the circuit are 5v each and have internal resistance of 0.2 ohms, then the reading in the ideal volt meter is??
-
UP 0 DOWN 0 0 13
13 Answers
1how 8.33 ??
okie... lemme try again...
(plz dun wait... i have to llogoff now,,,, if anyone else solve it now,then perffectly fine... or else i may post it in the evning if i can solve)
1oops yeah found out my mistake...
sorry :(
i gave the effective int resistance...
13hey chazi... i'm getting the answer as 7.94...
8.33 will be the answer if v cancel out d internal resistances of d batteries which are connected oppositely in d circuit, which is not d correct method...
well, i'm not sure of my answer... but i think dis is correct...!!!
13yeah... initially i got 7.5 and later 7.94...
i'll try it again... shayad is baar 8.33 aa jaye... [6] [4]
1
Well it took me some time to do it..
We need to find VA-VB
In the box D .. the resistance will become 0.4 and the potentiall difference in 0.
In box C.. The resistance is 0.1 and the potential difference is 0(using kirchoff).
So the total potential is 25 and the total resistance is 1.5..
So the current flowing is 25/1.5 =16.6
So the current flows to the battery E .. So the potential difference formed is (5+(16.6 x 0.2)) = 8.33
1arey bhayya!! siddharth... I have a conceptual doubt......
this is the resultant circuit . right??
voltage of the resultant cell is 25 volts...
and the resultant resistance is 1.5 ohms
since the volt meter is ideal the entire current generated in the resultant cell flows through the other cell
so its just like the two cells connected in series..........
so my opinion is that it is not correct to consider the current flowing through the individual cell .............
did u understand my doubt??
1yes i understood....
Check my diagram.
We have to find VA-VB... The cell is between A and B..
And dont forget that the cell E has some internal resistance. .So some potential is produced due to that resistance.. We need to consider that resistance too...
So we need to first consider the current flowing through the cell(ur diagram) and that current combines with the resistance in cell E to form another potential.