1please help
IN THE CIRCUIT SHOWN BELOW Va AND Vb be the potentials at A and B , R is the equivalent resistence between A and B , S1 and S2 are switches and DIODES are ideal 
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13 Answers
190.) not in syllabus.
but stil.........
apply formula q=q0(1-e-t/Γ)
differentiate it to get ' i' i.e. V/R.
176.) net resistance = 2r.r/2r+r = 2r/3
so, l' = (2r/3)/r X 100 > 50cm.
option b //
11
76.)
At balanced condition no current flows thro' cell C (whose emf is to be measured).
So,the internal resistance of cell C plays no role.
ε1 / ε2 = l1 / l2
\Rightarrow (ε/2) / (ε) = l / 100
\Rightarrow l = 50 cm
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R net = r (of cell D) + RAB
I = [ ε / R net ] = [ ε / (r + RAB) ]
VAB = I.RAB = [ ε.RAB / (r + RAB) ]
The balance pt. will be obtained if the potential of the potentiometer wire is sufficient enough to balance the emf of the cell (unknown) ,
i.e. VAB > (ε/2)
\Rightarrow [ ε.RAB / (r + RAB) ] > (ε/2)
\Rightarrow RAB > r
Hence, (A) and (C) are correct.
1ε1 / ε2 = l1 / l2
(ε/2) / (ε) = l / 100
l = 50 cm
[7]
first u told it plays no role... then u had put e/2 ... i din get...
and ans is B and C.
11I said
" ...the internal resistance of cell C plays no role. " not, "the cell plays no role."
190) time constatn equal to RC . so draw a line so at given time check which has more voltage , which ever has that one has max resistance
1Q) 76)
Let σ b the resistance per unit lenght..of AB
In balaced state I wll nit flow thru the lower circuit....
there fore.....I will flow only in the upper circuit...
I=E/(r+100σ)
potenrtial drop across AJ=I*σl
this shud be equal to E/2
so,I*σl=E/2
solving we get,..
σl/(r+100σ)=0.5
also obviously c is also r8
or,l=50+ r/(2σ)
so, l>50...
(b) is right