According to Ohm's law, the current flowing through a conductor is directly proprtional to the potential difference applied across its ends. So if the pd is 2 V then the charge flowing through the conductor at any Cross section of it should be equal to 2 Couloumb per second or 2 Ampere.
2 Answers
When two quantities are directly proportional, a "constant of proportionality" is required to form an equality. It is true that according to Ohm's Law, the current thru a conductor is directly proportional to the PD across. The constant of proportionality in this case would be (R), the resistance of the conductor in ohms (symbol: Greek letter Omega). Only when R = 1 ohm would the current thru the conductor be 2 A. For all other values of R, the current would be (V / R).
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$\begingroup$ Thanks but can you please like explain briefly what a constant of proportionality is and why why it is dropping over here in Ohm's equation out of nowhere. 😅😅 $\endgroup$ Feb 11, 2020 at 11:48
Consider the linear equation: y = 3 x. The slope of this line is 3. Mathematically, we can say that y ~ x. [y is proportional to x]. If we wish to express the proportionality as an equation, we need to insert a "constant of proportionality". In this specific case, the constant of proportionality is the slope, (3). Now we can write the equation as: y = 3 x.
In the case of Ohm's Law, (I) ~ V [NOTE: this was the original format, as Ohm did it. We have switched it around to V ~ (I) because it's easier to manipulate (V) in the lab.] Applying the requirement mentioned above, to make an equation out of V ~ (I); we need a constant of proportionality. It so happens that the constant in this case is (R); the resistance of the material being used. Ohm's Law then is rewritten as: V = R (I). Notice that this differs from the normal way we write Ohm's Law [V = (I) R]. What caused this alteration? Historically, Ohm showed that [delta V / delta (I) = slope = R].
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$\begingroup$ Thanks this helped.... A lot. I'm a 10th grade kid... I dont know who are but still thanks a lot 😁 $\endgroup$ Feb 11, 2020 at 13:35