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Q.

When a resistor of 11 ohm is connected in series with an electric cell the current flowing in it is 0.5 A. Instead, when a resistor of 5Ωis connected to the same electric cell in series, the current increases by 0.4A. The internal resistance of cell is

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a

1.5Ω

b

2Ω

c

2.5Ω

d

3.5Ω

answer is B.

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Detailed Solution

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Current taken from the cell.
i=Er+R;0.5=E11+r→(1)0.5+0.4=E5+r→(2);(1)(2);59=5+r11+r Solving r=2.5Ω

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