A potentiometer circuit is set up as shown. T | vedclass

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(B) When the two-way key is switched off,the current flowing through resistors $R$ and $X$ is $I = 1.0 \, A$.When the key between terminals $1$ and $2

Vedclass · Accepted Answer

$kl_1 \, \Omega, k(l_2 - l_1) \, \Omega$

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(B) When the two-way key is switched off,the current flowing through resistors $R$ and $X$ is $I = 1.0 \, A$.When the key between terminals $1$ and $2$ is plugged in,the potential difference across resistor $R$ is balanced by the potentiometer wire of length $l_1$:$V_R = I \cdot R = k \cdot l_1$Since $I = 1 \, A$,we have $R = k \cdot l_1 \, \Omega$.When the key between terminals $1$ and $3$ is plugged in,the potential difference across the combination of resistors $(R + X)$ is balanced by the potentiometer wire of length $l_2$:$V_{R+X} = I \cdot (R + X) = k \cdot l_2$Since $I = 1 \, A$,we have $R + X = k \cdot l_2 \, \Omega$.Substituting $R = k \cdot l_1$ into the equation for $(R + X)$:$k \cdot l_1 + X = k \cdot l_2$$X = k(l_2 - l_1) \, \Omega$.Thus,the magnitudes of resistors $R$ and $X$ are $kl_1 \, \Omega$ and $k(l_2 - l_1) \, \Omega$ respectively.

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