DMUKH ELECTRICITY 1 pls explain also

90.) not in syllabus.

but stil.........

apply formula q=q₀(1-e^-t/Γ)

differentiate it to get ' i' i.e. V/R.

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) + R_AB

I = [ ε / R _net ] = [ ε / (r + R_AB) ]

V_AB = I.R_AB = [ ε.R_AB / (r + R_AB) ]

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. V_AB > (ε/2)

\Rightarrow [ ε.R_AB / (r + R_AB) ] > (ε/2)

\Rightarrow R_AB > r

Hence, (A) and (C) are correct.

I said
" ...the internal resistance of cell C plays no role. " not, "the cell plays no role."

for 76 ANSWERS ARE B AND C [50]