CURRENT ELECTRICITY
Ans. Principle of potentiometer:- The potential difference across any portion of the wire is directly proportional to the length of that portion.
$$V= {\Phi l}$$
Comparing emf's of two cells using potentiometer: Consider two cells connected to potentiometer as shown in the circuit diagram and cells e.m.f are $E_{1},E_{2}$ ·
According to kirchhoff's second law
$$\begin{array}{c}-\:E_1\:+\:0\:+\:\Phi\:l_1\:=\:0\\-\:E_2\:+\:0\:+\:\Phi\:l_2\:=\:0\end{array}$$
From the above equations $$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$$
Ans. Principle of potentiometer:- The potential difference across any portion of. the wire is directly proportional to the length of that portion..
$$V={\Phi l}$$
Determination of Internal resistance of a cell using potentiometer: Consider a cell and high resistance connected to potentiometer as shown in the circuit diagram and cell e.m.f. is E, when resistance is connected to circuit, voltage is 'V' therefore,
according to kirchhoff's second law
$$\begin{array}{cc}
&-\:E\:+\:0\:+\:\phi\:l_{1}\:=\:0 \\
& -\:V\:+\:0\:+\:\phi\:l_{2}\:=\:0
\end{array}\\$$
From the above equations $$\frac{E}{V}=\frac{l_{1}}{l_{2}}\quad-----(1)$$
But, $E$ = $i( r$ + $R)$ and $V=iR$
From eq (1) $$\left(\frac{r+R}{R}\right)=\frac{l_{1}}{l_{2}}$$
Therefore internal resistance of a cell $$r=R\left(\frac{l_{1}}{l_{2}}-1\right)$$