A constant voltage V₀ is applied to a potentiometer of resistance R connected to an ammeter. A constant resistor r is connected to an ammeter. A constant resistor r is connected to the sliding contact of the potentiometer and the fixed end of the potentiometer. If the reading of ammeter vary as the sliding contact is moved from one end of the potentiometer to the other. The resistance of ammeter is assumed to be negligible. Find the minimum reading of the Ammeter.

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$\frac{V_{0} (R + r)}{Rr}$

$\frac{V_{0}}{R + r}$

$\frac{V_{0} R}{r (r + R / 4)}$

$\frac{V_{0} r}{R (r + R / 4)}$

answer is D.

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Detailed Solution

When the jockey is at J, let the resistances of the potentiometer wire on either side of it are $x, R - x$

Resistance of the system

$R_{e} = \frac{rx}{r + x} + R - x \Rightarrow R_{e} = \frac{R (r + x) - x^{2}}{r + x}$

The current drawn from the battery is $I = \frac{V_{0}}{R_{e}}$

$I = \frac{V_{0} (r + x)}{Rr + R x - x^{2}}$

current through ammeter is $I_{1} = \frac{Ir}{r + x} = \frac{V_{0} r}{Rr + R x - x^{2}}$

This is the reading of ammeter.

This is minimum when $\frac{{dI}_{1}}{d x} = 0$

$\Rightarrow - V_{0} r [\frac{R - 2 x}{{(Rr + R x - x^{2})}^{2}}] = 0$

$\Rightarrow R - 2 x = 0 \Rightarrow x = R / 2$

$\Rightarrow∴ {(I_{1})}_{min} = \frac{V_{0} r}{Rr + R \frac{R}{2} - {(\frac{R}{2})}^{2}} = \frac{V_{0} r}{R (r + \frac{R}{4})} .$

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