A new flashlight cell of emf 1.5 volts gives a current of 15 amps, when connected directly to an ammeter of resistance 0.04 ohm . The internal resistance of the cell is
A
0.04Ω
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B
0.06Ω
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C
0.10Ω
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D
10Ω
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Solution
The correct option is B0.06Ω By using i=ER+r⇒15=1.50.04+r⇒r=0.06Ω