The figure shows a potentiometer wire AB havi | vedclass

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(A) $1$. First, calculate the current $I$ flowing through the potentiometer wire $AB$ in the primary circuit:$I = \frac{E}{R_{total}} = \frac{5 	ext{

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$8$

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(A) $1$. First, calculate the current $I$ flowing through the potentiometer wire $AB$ in the primary circuit:$I = \frac{E}{R_{total}} = \frac{5 	ext{ V}}{45 	ext{ } \Omega + 5 	ext{ } \Omega} = \frac{5}{50} 	ext{ A} = 0.1 	ext{ A}$.$2$. Calculate the potential drop across the potentiometer wire $AB$:$V_{AB} = I 	imes R_{AB} = 0.1 	ext{ A} 	imes 5 	ext{ } \Omega = 0.5 	ext{ V}$.$3$. The potential gradient $k$ along the wire is given by:$k = \frac{V_{AB}}{L} = \frac{0.5 	ext{ V}}{10 	ext{ m}} = 0.05 	ext{ V/m}$.$4$. For the balancing length $AP = l$, the potential drop across $AP$ must equal the e.m.f. of the secondary cell $(E' = 0.4 	ext{ V})$:$V_{AP} = k 	imes l = 0.4 	ext{ V}$.$0.05 	ext{ V/m} 	imes l = 0.4 	ext{ V}$.$l = \frac{0.4}{0.05} 	ext{ m} = 8 	ext{ m}$.Therefore, the balancing length $AP$ is $8 	ext{ m}$.

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