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When we increase the area of cross section of a wire, we find that with same voltage provided there is a twice amount of current flowing through resistor than before.

Either you think that " this is because of resistance inversely proportional to area of resistor" or anything else, give a reason for why that happens.

If your answer is same as given in quote then, give reason for why increasing area decreases resistance.

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A simple intuition is that when we increase the area of cross section, more number of electrons can pass through the area as compared to initial wire. More electrons results in more charge crossing the cross section, thus more current.

Now let's look at it with some Math:

Resistance of a wire is given by : $$R = \rho \frac{l}{A}$$ $R$ = resistance of the wire
$\rho$ = resistivity of the wire (depends upon material of the wire)
$l$ = length of the wire
$A$ = Area of cross section


If we take wire of same length and of same material i.e. same resistivity then: $$R\propto \frac{1}{A}$$

Thus if you increase the area of cross section then resistance of the wire (keeping length and resistivity same) will decrease.
Also from Ohm's Law ( $V=iR$ ) we can say that current will increase if potential difference across the wire is same in both the cases.

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  • $\begingroup$ Seems similar to my answer but yours is not so explained. $\endgroup$
    – user316791
    Dec 23, 2021 at 9:25
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Not a professional.

This is because the number of electric field lines passing through the resistor increaaes as you increase area of cross section. Therefore, the double number of electrons get the same amount of force as it was before(for initial area) to move through and hence double current

Actually to imagine that number of field lines you can set a fixed space between lines and specify the strength of that field density. Now when increase the area, you see the density of field lines remains same.

That's the key point.

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