Question 604: Current Electricity

Question

A potentiometer wire has length $L$ For given cell of emf $E$, the balancing length is $\frac{L}{3}$ from the positive end of the wire. If the length of potentiometer wire is increased by $50 \%$, then for the same cell, the balance point is obtained at length

Accepted Answer

For cell of emf $E$ balancing length $=\frac{L}{3}=\frac{1}{3} 	imes L($ length of wire $)$
Increased length of wire $=L+\frac{L}{2}=\frac{3 L}{2}$
Since, the cell is the same, therefore the new balancing length $=\frac{1}{3}$ (new increased length)
$$\begin{aligned}
& =\frac{1}{3} 	imes \frac{3}{2} L \
& =\frac{L}{2}
\end{aligned}$$
Therefore, for the same cell the balance point will be as length $\frac{L}{2}$ from positive end.

Answer

$\frac{L}{2}$ from positive end

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$\frac{L}{5}$ from positive end

Answer

$\frac{L}{3}$ from positive end

Answer

$\frac{L}{4}$ from positive end

Current Electricity

A potentiometer wire has length L For given cell of emf E, the balancing length is \frac{L}{3} from the positive end of the wire. If the length of potentiometer wire is increased by 50 \%, then for the same cell, the balance point is obtained at length

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