Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let, the internal resistance be $r$ and the resistance adjusted in rheostat be $R$.

Then, the circuit is represented for the given situation as follows,



$V=E-Ir...\left(1\right)$

$I=\frac{E}{r+R}...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$V=I(r+R)-Ir$

$\Rightarrow V=IR$

$\Rightarrow V=\frac{ER}{r+R}...\left(3\right)$

When $R=0$ then $V=0$

And when $R=\infty$ then $V=\frac{E}{\frac{r}{\infty }+1}$

i.e. $V=E$

Plotting graph from equation $\left(3\right)$, we get


Hence, option (d) is the correct option.