Question

(a) State the working principle of potentiometer. Explain with the help of a circuit diagram, how the potentiometer is used to determine the internal resistance of the given primary cell.
(b) A battery of emf $10V$ and internal resistance $3\Omega$ is connected to a resistor $R$. If the current in the circuit is $0.5A$, calculate the value of $R$.

Answer 1

Working principle : Potential difference across length of potentiometer wire is proportional to the length of the wire.
$E=kL$ where, $E$ is the emf and $k$ is the potential gradient.
In the figure:
Cell emf, $e$ whose internal resistance $r$ is to be determined is connected via resistance box R.B. through key $K_2$.

This key when open, balance is obtained at $l_1\left(A{N}_1\right)$ .
So, $e=k{l}_1$
If $v$ is terminal potential difference, balance is obtained at length $l_2\left(A{N}_2\right)$.
So, $v=k{l}_2$
Now, $e=I(R+r)$

where, $I$ is the current and $v=IR$

Dividing the equations.
$\frac{e}{v}=\frac{l_1}{l_2}=\frac{R+r}{r}$

$\Rightarrow r=R(\frac{l_1}{l_2}-1)$
EMF, $V=10v$
Current, $I=0.5A$
Internal resistance, $r=3\Omega$
Resistance of resistor be $R$
$I=\frac{V}{R+3}$

$R+3=20$

$R=17$ $\Omega$