A battery of emf 2 volt and internal resistance 0.1 ohm is being charged with a current of 5 ampere. The potential difference between the two terminals of the batteries is
A
2V
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B
0.5V
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C
1.5V
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D
2.5V
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Solution
The correct option is D2.5V When a cell is being charged the potential drop across it is given by V=E+i×Rint Hence , V=2+0.1×5 volts