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A conductor of resistance R is connected to a battery. A charge of 36 coulomb passed through the conductor per 2min. Then p.d. across the end of the wire is 4 V. Then the value of ‘R’ will be

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12.3 $Ω$

13.3 $Ω$

11.3 $Ω$

10.3 $Ω$

answer is D.

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Detailed Solution

$R = \frac{V}{I} = \frac{Vt}{Q}; R = \frac{4 \times 2 \times 60}{36} = 13.3 Ω$

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