When a 6 Ω resistor is connected across the terminals of a 3 V battery, the number of coulombs passing through the resistor per second is

Given that - R = 6 ohm

V = 3 volt

n = ?

so by ohms law , V = I X R

I = V \ R = 3 \ 6 = 0.5 A

we know that I = Q / T here t is 1 sec

Q = I X T = 0.5 X 1 = 0.5 C

By the law of quantization ,

Q = n x e

n = Q / e = 0.5 / 1.6 x 10 ^-19

n = 0.3125 x 10^19

hoped so u got the answer mark me as brainellist :)

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