The emf of a Daniel cell is $1 .08 V$ When the terminals of the cells are connected to a resistance of $3 Ω$ , the potential difference across the terminals is found to be $0 .6 V$ . Then the internal resistance of the cell is

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$1 .8 Ω$

$0 .2 Ω$

$2 .4 Ω$

$3 .24 Ω$

answer is B.

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Detailed Solution

emf of a Daniel cell $E = 1.08 V$ Resistance $R = 3 Ω$
Potential difference $V = 0.6 V$ Internal resistance $r$

$V = E - Ir$

$\Rightarrow 0.6 = 1.08 - Ir \Rightarrow Ir = 0.48$

$(\frac{1.08}{R + r}) r = 0.48 \Rightarrow r = 2.4 Ω$

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