The total length of the potentiometer wire AB | vedclass

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(B) The given circuit represents a Wheatstone bridge configuration where the galvanometer shows zero deflection at point $P$. This implies that the br

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$30$

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(B) The given circuit represents a Wheatstone bridge configuration where the galvanometer shows zero deflection at point $P$. This implies that the bridge is balanced. For a balanced Wheatstone bridge, the ratio of resistances in the arms must be equal: $\frac{R_1}{R_2} = \frac{R_{AP}}{R_{PB}}$ Given $R_1 = 6 \ \Omega$ and $R_2 = 4 \ \Omega$. Let $L$ be the total length of the wire $AB = 50 	ext{ cm}$. Let $\ell_{AP} = x$ and $\ell_{PB} = 50 - x$. Since the resistance of the wire is proportional to its length $(R = ho \frac{\ell}{A})$, we have: $\frac{6}{4} = \frac{x}{50 - x}$ $\frac{3}{2} = \frac{x}{50 - x}$ $3(50 - x) = 2x$ $150 - 3x = 2x$ $5x = 150$ $x = 30 	ext{ cm}$. Therefore, the length $AP$ is $30 	ext{ cm}$.

The total length of the potentiometer wire $AB$ is $50 \text{ cm}$ in the arrangement as shown in the figure. If $P$ is the point where the galvanometer shows zero reading, then the length $AP$ is . . . . . . $\text{ cm}$.

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