The emf of a storage battery is $90 V$ before charging and $100 V$ after charging. When charging began the current was $10 A$ . What is the current at the end of charging if the internal resistance of the storage battery during the whole process of charging may be taken as constant and equal to 2 $Ω$ ?

5 A

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10 A

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20 A

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40 A

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Solution

The correct option is A $5 A$
Here emf is same before charging and after charging as $E$ . Taking initial voltage as $V_{i}$ and final voltage as $V_{f}$ .
$V_{i} = E - i_{i} . r E = V_{i} + i_{i} . r = 90 + 10 \times 2 = 110 v o l t s$

Let $i_{f}$ be the current at the end of charging.
Then, $V_{f} = E - i_{f} . r$
or,
$i_{f} = \frac{E - V_{f}}{r} = \frac{110 - 100}{2}$
$i_{f} = 5 A$

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