Question

When a 4 Ω resistor is connected across the terminals of a 2 V battery, the number of coulombs passing through the resistor per second is:

Answer 1

Answer:

Given that - R = 4 ohm

V = 2 volt

n = ?

so by ohms law , V = I X R

I = V / R = 2 / 4 = 1 / 2 = 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

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Answer 2

Given : A 4 Ω resistor is connected across the terminals of a 2 V battery.

To find : The number of coulombs passing through the resistor per second.

Solution : we know, current is the amount of charge passing through a resistor per second.

means, Indirectly question wants us to find current passing through the resistor.

from Ohm's law,

I = V/R = (2 volts)/(4 Ω) = 0.5 A = 0.5 C/s [ as we know, 1A = 1C/s ]

Therefore 0.5 Coulombs passing through the resistor per second.

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