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(B) Let the potential at $B$ be $V_B = 0\,V$. Since the $6\,V$ battery is connected across $AB$,the potential at $A$ is $V_A = 6\,V$.The potential at

Vedclass · Accepted Answer

$67$

Vedclass · Answer

(B) Let the potential at $B$ be $V_B = 0\,V$. Since the $6\,V$ battery is connected across $AB$,the potential at $A$ is $V_A = 6\,V$.The potential at any point $x$ (in $cm$) from $A$ on the wire $AB$ is given by $V(x) = V_A - (V_A - V_B) \cdot \frac{x}{L} = 6 - 6 \cdot \frac{x}{100} = 6(1 - 0.01x)$.The potential at point $C$ is determined by the $4\,V$ battery. Since no current flows through the branch containing the $4\,V$ battery and the $1\,\Omega$ resistor (as it is an open circuit at $C$),the potential at $C$ is equal to the potential of the positive terminal of the $4\,V$ battery relative to $A$.Specifically,$V_C = V_A - 4\,V = 6\,V - 4\,V = 2\,V$.We want to find the point $D$ at distance $x$ from $A$ such that $V(D) = V_C = 2\,V$.Setting $6(1 - 0.01x) = 2$,we get $1 - 0.01x = \frac{2}{6} = \frac{1}{3}$.$0.01x = 1 - \frac{1}{3} = \frac{2}{3}$.$x = \frac{2}{3} \cdot 100 = 66.67\,cm$.Rounding to the nearest integer,we get $x \approx 67\,cm$.

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