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The figure shows the diagram of a potentiometer.

1. Give the principle of a potentiometer. 

2. The length of AB is 3m and resistance per unit length of the potentiometer wire is 4Ω/m. If E1 = 4V, R = 20Ω and E2 = 1V find the length of the potentiometer wire that balance E2.

3. If E2>E1 can we get the null deflection in galvanometer. Give reason.

1 Answer

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1. When a constant current is flowing through a wire having uniform area of cross section and uniform composition the potential difference across any length of the wire is directly proportional to its length.
V ∝ l.

2. The resistance of potentiometer

Rp = 4 × 3 = 12Ω

Current through the potentiometer

I = \(\frac{E}{R+R_p}=\frac{4}{20+12}\) = 0.125 A

Potential across potentiometer wire

VAB = IRp = 0.125 × 12 = 1.5V

Potential gradient k = \(\frac{V_{AB}}{I}=\frac{1.5}{3}\) = 0.5 V/m

Balancing length for cell E2 is given from equation
E2 = kI2 = 0.5I2

\(I_2=\frac{E}{k}=\frac{1}{0.5}\) = 2 m

3. If E2 > E1 , we will not get null deflection. The potential difference across the potentiometer wire AB should be higher than the emf of E2 .

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