A cell of emf E and internal resistance r is connected across a v… - Vidyadip

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 $(i)$ resistance $R$ and $(ii)$ current $I$.
It is found that when $R=4 \Omega$, the current is $1 A$ and when $R$ is increased to $9 \Omega$, the current reduces to $0.5 A$ .Find the values of the emf $E$ and internal resistance $r$.

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Answer

Internal resistance usually means the electrical resistance inside batteries and power supplies that can limit the potential difference that can be supplied to an external load
$\because V=\left(\frac{E}{R+r}\right) R=\frac{E}{1+r / R}$
$\Rightarrow$ with the increase of $R , V$ increases
Graph between terminal voltage $(V)$ and Current $(I)$

When $R=4 \Omega$ and $I =1 A$.
We know that, terminal voltage $, V = E - Ir.$
$\Rightarrow V=I R=4=E-I r$
$\Rightarrow E-r=4 \ldots \text { (i) }$
When $R=9 \Omega$ and $I =0.5 A$, then
$V=I R=0.5 \times 9=E-0.5 r$
$\Rightarrow E-05 r=4.5 \ldots(i i)$
On solving Eqs. $(i)$ and $(ii),$ we get
$r=1 \Omega$ and $E =5 V$
So from the above calculation, it was found that the internal resistance of the cell is $1 \text { ohm}$ and emf is $5$ volt.

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