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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 $\left(\mathrm{E}_{1}>\mathrm{E}_{2}\right)$
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$2: 3$
\(\frac{E_{a b}}{E_{a c}}=\frac{L_{a b}}{L_{a c}}=\frac{E_{1}}{E_{1}-E_{2}}=\frac{0.9}{0.3}=3\)
\(E_{1}=3 E_{1}-3 E_{2}\)
\(2 E_{1}=3 E_{2}\)
\(\frac{E_{1}}{E_{2}}=\frac{3}{2}\)
\(\therefore \frac{E_{2}}{E_{1}}=\frac{2}{3}\)
\(E_{1}=3 E_{1}-3 E_{2}\)
\(2 E_{1}=3 E_{2}\)
\(\frac{E_{1}}{E_{2}}=\frac{3}{2}\)
\(\therefore \frac{E_{2}}{E_{1}}=\frac{2}{3}\)
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