A potentiometer wire $A B$ of length 600 cm and resistance $R=15 r$ is connected to a cell of emf $\

Question

A potentiometer wire $A B$ of length 600 cm and resistance $R=15 r$ is connected to a cell of emf $\mathcal{E}$ and internal resistance $r$. A secondary circuit containing a cell of emf $\mathcal{E} / 2$ and internal resistance $r$ is connected to the potentiometer as shown. At what distance $x$ from point $A$ should the jockey touch the wire to get zero deflection in the galvanometer?

Professor PC Thomas Classes · Answer

Primary Circuit Current (I): The total resistance of the primary circuit is $R_{	ext {total }}=R_{	ext {wire }}+r=15 r+r=16 r$. $$ I=\frac{\mathcal{E}}{16 r} $$ Potential Gradient ( $k$ ): The potential drop across the wire $A B$ is $V_{A B}=I 	imes R_{	ext {wire }}=\frac{\mathcal{E}}{16 r} 	imes 15 r=\frac{15 \mathcal{E}}{16}$. Potential gradient $k=\frac{V_{A B}}{L}=\frac{15 \mathcal{E}}{16 	imes 600}=\frac{\mathcal{E}}{640} \mathrm{~V} / \mathrm{cm}$. \Balancing Condition: For zero deflection, the potential drop across length $x$ must equal the emf of the secondary cell $(\mathcal{E} / 2)$. Note: Since there is no current in the secondary circuit at balance, the internal resistance $r$ does not cause a potential drop. 
$ \begin{gathered} V_x=\frac{\mathcal{E}}{2} \ k \cdot x=\frac{\mathcal{E}}{2} \ \frac{\mathcal{E}}{640} \cdot x=\frac{\mathcal{E}}{2} \end{gathered} $
$ x=\frac{640}{2}=320 \mathrm{~cm} $