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

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V=I R=(E R/R+r) ⇒ V=(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=∞ so the graph can be as drawn below.

Q. 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

$V = I R = \frac{ER}{R + r} \Rightarrow V = \frac{E}{1 + r / R}$

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.

Q. 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

V=IR=R+rER​⇒V=1+r/RE​ So as R increases, V also increases. Also at R=0,V=0. But maximum value of V can be E for R=∞ so the graph can be as drawn below.

Q. 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

$V = I R = \frac{ER}{R + r} \Rightarrow V = \frac{E}{1 + r / R}$

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.