First slide

Ohm's law

Question

A cell of emf E has an internal resistance r and is connected to a rheostat. When resistance R of rheostat is changed, correct graph of potential difference across it is

Moderate

A
B
C
D

Solution

$V = IR = \frac{ER}{R + r} \Rightarrow V = \frac{E}{1 + r / R}$
So as R increases, V also increases. Also at R = 0, V = 0. But maximum value of V can be E for $R = \infty$ so the graph can be as drawn below.

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