A potentiometer is connected between A and B | vedclass

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Question

(A) Let $V_{AB}$ be the potential difference between $A$ and $B$ and $V_{BC}$ be the potential difference between $B$ and $C$. Let $k$ be the potentia

Vedclass · Accepted Answer

$179$

Vedclass · Answer

(A) Let $V_{AB}$ be the potential difference between $A$ and $B$ and $V_{BC}$ be the potential difference between $B$ and $C$. Let $k$ be the potential gradient of the potentiometer wire.When connected between $A$ and $B$,the balance length $l_1 = 203.6 \, cm$. Thus,$V_{AB} = k \cdot l_1 = k \cdot 203.6 \, cm$. When the end is shifted to $C$,the total potential difference is $V_{AC} = V_{AB} + V_{BC}$. The balance length is $l_2 = 24.6 \, cm$. However,since $V_{AC} > V_{AB}$,the balance length should increase. The problem implies the polarity is reversed or the connection is $V_{AB} - V_{BC}$. Given the balance point is $24.6 \, cm$,we have $V_{AB} - V_{BC} = k \cdot 24.6 \, cm$.Subtracting the two equations:$V_{AB} = k \cdot 203.6 \, cm$$V_{AB} - V_{BC} = k \cdot 24.6 \, cm$$V_{BC} = k \cdot (203.6 - 24.6) = k \cdot 179.0 \, cm$.Thus,the balance point for $B$ and $C$ is $179.0 \, cm$.

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