A potentiometer wire of length 10 m carries a steady current. EMF $\varepsilon_1$ balances at 6 m; the same cell with an external 4 Ω resistor across it balances at 4 m. The internal resistance $r$ of the cell is:
A3 Ω
B8 Ω
C1 Ω
D2 Ω
Answer & Solution
Correct answer: D. 2 Ω
Potentiometer with cell + external R measures terminal voltage $V = \varepsilon R/(R+r)$. So $V/\varepsilon = \ell_2/\ell_1$ ⇒ $R/(R+r) = 4/6 = 2/3$ ⇒ $3R = 2R + 2r$ ⇒ $r = R/2 = 2\ \Omega$.
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