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Q.

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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a

1.8Ω

b

0.2Ω

c

2.4Ω

d

3.24Ω

answer is B.

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Detailed Solution

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

V=E-Ir

⇒0.6=1.08-Ir ⇒Ir=0.48

1.08R+rr=0.48⇒r=2.4Ω

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