Question

Consider the two cells having emf $E_{1}$ and $E_{2}\left(E_{1}>E_{2}\right)$ connected as shown in the figure. A potentiometer is used to measure potential difference between $P$ and $Q$ and the balancing length of the potentiometer wire is $0.8\, m$. Same potentiometer is then used to measure potential difference between $P$ and $R$ and the balancing length is $0.2\, m$. Then, the ratio $E_{1} / E_{2}$ is

• A. 44654
• B. 44685
• C. 44684
• D. 44652

Answer 1

The Correct Option is A

Consider, the diagram shown below
$\left(V_{p}-V_{\theta}\right)=E_{1} \propto L_{1}$ ... (i)
$V_{p}-V_{R}=\left(V_{P}-V_{Q}\right)+\left(V_{Q}-V_{R}\right)$
$=(E_{1}+\left(-E_{2}\right)=E_{1}-E_{2} \propto L_{2}$ ...(ii)
Where. $L_{1}$ And $L_{2}$ are the lengths of potentiometer
From Eqs. (i) And (ii), we get
$\frac{E_{1}}{E_{1}-E_{2}}=\frac{L_{1}}{L_{2}}=\frac{0.8}{0.2}$
$\Rightarrow \frac{E_{1}}{E_{1}-E_{2}}=\frac{8}{2}=4$
$\Rightarrow E_{1}=4 E_{1}-4 E_{2}$
$\Rightarrow 4 E_{2}=3 E_{1}$
$\Rightarrow \frac{E_{1}}{E_{2}}=\frac{4}{3}$