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In a potentiometer experiment the wire is 10 m long and has a resistance of 20 Ω. It is connected to a driver battery of emf 4 V (negligible internal…

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In a potentiometer experiment the wire is

10\ \mathrm{m}

long and has a resistance of

20\ \Omega

. It is connected to a driver battery of emf

4\ \mathrm{V}

(negligible internal resistance) in series with a resistor of

20\ \Omega

. A cell of unknown internal resistance balances at

7.5\ \mathrm{m}

on open circuit. When an

8\ \Omega

resistor is connected across the cell, the balance length becomes

6.0\ \mathrm{m}

. The internal resistance of the cell is

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Answer: (A) $2\ \Omega$

The current in the potentiometer circuit is

4/(20+20)=0.1\ \mathrm{A}

, so the potential gradient is

0.1\times20/10 = 0.2\ \mathrm{V\,m^{-1}}

. Open-circuit emf:

E = 0.2\times7.5=1.5\ \mathrm{V}

. Terminal voltage with the load:

V=0.2\times6.0=1.2\ \mathrm{V}

. For the cell,

V = E\dfrac{R}{R+r}

, so

r = R\left(\dfrac EV-1\right)=8\left(\dfrac{1.5}{1.2}-1\right)=2\ \Omega

. Equivalently

r=R(\ell_1-\ell_2)/\ell_2 = 8(1.5)/6

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