A potentiometer is used to measure the potential difference between $\mathrm{A}$ and $\mathrm{B}$, the null point is obtained at $0.9 \mathrm{~m}$. Now potential difference between $\mathrm{A}$ and $\mathrm{C}$ is measured, the null point is obtained at $0 \cdot 3 \mathrm{~m}$. The ratio…

Question

A potentiometer is used to measure the potential difference between $\mathrm{A}$ and $\mathrm{B}$, the null point is obtained at $0.9 \mathrm{~m}$. Now potential difference between $\mathrm{A}$ and $\mathrm{C}$ is measured, the null point is obtained at $0 \cdot 3 \mathrm{~m}$. The ratio $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}$ is $\left(\mathrm{E}_{1}>\mathrm{E}_{2}\right)$

Aicharya · Accepted Answer

$\frac{E_{a b}}{E_{a c}}=\frac{L_{a b}}{L_{a c}}=\frac{E_{1}}{E_{1}-E_{2}}=\frac{0.9}{0.3}=3$ $E_{1}=3 E_{1}-3 E_{2}$ $2 E_{1}=3 E_{2}$ $\frac{E_{1}}{E_{2}}=\frac{3}{2}$ $\therefore \frac{E_{2}}{E_{1}}=\frac{2}{3}$

A potentiometer is used to measure the potential difference between $\mathrm{A}$ and $\mathrm{B}$, the null…

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