Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance of 0.015 are joined in series to provide a supply to a resistance of 8.5 . What are the current drawn from the supply and it's terminal voltage?

11.8749

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12.8749

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13.1749

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10.5749

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Solution

The correct option is A $11.8749$
$V T = V - i r V T = 12 - 1.39 \times 0.09 V T = 11.8749$

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Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance of 0.015 are joined in series to provide a supply to a resistance of 8.5 . What are the current drawn from the supply and it's terminal voltage?

The correct option is A 11.8749VT=V−irVT=12−1.39×0.09VT=11.8749

The correct option is A $11.8749$
$V T = V - i r V T = 12 - 1.39 \times 0.09 V T = 11.8749$