Question

A cell of emf 'E' and internal resistance 'r' is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current i.
It is found that when $R=4\Omega$, the current is $1A$ when R is increased to $9\Omega$, the current reduces to $0.5A$. Find the values of the emf E and internal resistance r

Answer 1

Here the terminal voltage , $V=E-ir...\left(1\right)$ and the current $i=E/(R+r)$

Now, $V=E-\frac{E}{(R+r)}r=\frac{ER}{R+r}=\frac{E}{(1+r/R)}=...\left(2\right)$

According to equation (1) and (2), the graph V vs i and V vs R will be as shown in figure.

As $i=E/(R+r)$ so $1=E/(4+r)$ or $E=4+r...\left(3\right)$ and

$0.5=E/(9+r)$ or $2E=9+r....\left(4\right)$

using (3), $2(4+r)=9+r$ or $r=1\Omega$

Thus, $E=4+1=5V$