A resistance of 4Ωand a wire of length 5 metres and resistance5Ωare joined in series and connected to a cell of emf 10 V and internal resistance1Ω. A parallel combination of two identical cells is balanced across 300 cm of the wire. The emf E of each cell is
A
1.5 V
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B
3.0 V
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C
0.67 V
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D
1.33 V
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Solution
The correct option is B3.0 V By using Eeq=(eR+Rh+r).RL×l⇒E=10(5+4+1)×55×3⇒E=3 V