A flashlight cell of emf. 1.5volt gives 15 ampere current when connected to an ammeter of resistance 0.04Ω. The internal resistance of the cell will be
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 The emf of cell is, E=1.5V The current in circuit is, I=15A The resistance of ammeter in circuit is, RA=0.04Ω The equation for current in circuit is I=Er+RA where, r is the internal resistance of cell. 15=1.5r+0.04 r+0.04=1.515=0.1 r=0.1−0.04=0.06Ω