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A uniform resistance wire AB of total resistance R₀ is connected in a circuit with an EMF source. A sliding contact at C divides the wire into two parts: AC = f*R₀ and CB = (1-f)*R₀, where 0 <= f <= 1. The contact C is also connected to another part of the circuit. For what value of f is the ammeter reading minimum?

1. (A) Minimum if f = 0
2. (B) Minimum if f = 1
3. (C) Minimum if f = 1/2
4. (D) Minimum if f = 1/3

Correct answer: (C) Minimum if f = 1/2

Solution

When the sliding contact C divides wire AB into AC = f*R₀ and CB = (1-f)*R₀, and both ends A and B are connected to the circuit while C is also connected (making the two segments parallel), the equivalent resistance is R_eq = (f*R₀ * (1-f)*R₀) / (f*R₀ + (1-f)*R₀) = f*(1-f)*R₀² / R₀ = f*(1-f)*R₀. The function f*(1-f) = f - f² has maximum at df/df = 1 - 2f = 0, giving f = 1/2, with maximum value 1/4. So R_eq is maximum at f = 1/2, and the ammeter current (I = E / (R_eq + other R)) is minimum at f = 1/2.

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