A potentiometer PQ is set up to compare two r | vedclass

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(D) Let $x$ be the potential gradient of the potentiometer wire $PQ$.When the key $K_3$ is plugged in between $2$ and $1$,the potential difference acr

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$\frac{l_1}{l_2 - l_1}$

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(D) Let $x$ be the potential gradient of the potentiometer wire $PQ$.When the key $K_3$ is plugged in between $2$ and $1$,the potential difference across $R_1$ is balanced:$V_1 = I R_1 = x l_1$When the key $K_3$ is plugged in between $3$ and $1$,the potential difference across the series combination of $R_1$ and $R_2$ is balanced:$V_2 = I (R_1 + R_2) = x l_2$Dividing the two equations:$\frac{I R_1}{I (R_1 + R_2)} = \frac{x l_1}{x l_2}$$\frac{R_1}{R_1 + R_2} = \frac{l_1}{l_2}$Inverting both sides:$\frac{R_1 + R_2}{R_1} = \frac{l_2}{l_1}$$1 + \frac{R_2}{R_1} = \frac{l_2}{l_1}$$\frac{R_2}{R_1} = \frac{l_2}{l_1} - 1 = \frac{l_2 - l_1}{l_1}$Therefore,the ratio $\frac{R_1}{R_2} = \frac{l_1}{l_2 - l_1}$.

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