A cell of emf E having an internal resistance r is connected to an external resistance R. The potential difference V across the resistance R varies with R as shown in the figure by the curve;

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Solution

The correct option is B B
The voltage drop across a cell having internal resistance is given by:
$V = i (R + r)$
$V_{R} = i \times R$
$V_{R} = (\frac{V}{R + r}) R$
case (i)
Where $R = 0$
$V_{R} = 0$
case (ii)
Where $R = large, (r << R)$
$V_{R} = \frac{V}{(R + r)} R R >> R$
$= \frac{V}{R} R$
$V_{R} = V$

Therefore, the graph $B$ represents the correct curve.

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A cell of emf E having an internal resistance r is connected to an external resistance R. The potential difference V across the resistance R varies with R as shown in the figure by the curve;

The correct option is B BThe voltage drop across a cell having internal resistance is given by:V=i(R+r)VR=i×RVR=(VR+r)Rcase (i)Where R=0VR=0case (ii)Where R=large,(r<<R)VR=V(R+r)RR>>R=VRRVR=VTherefore, the graph B represents the correct curve.