4 added latex for readability edit approved May 9 '16 at 1:56 GodotMisogi 63977 silver badges1414 bronze badges Let us consider a fixed voltage. I = nAve$$I = nAve$$ is the relation between current, number of free electrons in unit volume (n)$$(n)$$, cross-sectional area A$$A$$ and electron drift velocity v$$v$$. (e$$e$$ is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A$$A$$ of the original wire. Neither the n$$n$$ quantity, nor v$$v$$ is changed since we did not change the material of the wire. Now how the ohm'sOhm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n$$n$$ is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? (Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision  , isn't it?) Let us consider a fixed voltage. I = nAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? (Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision  , isn't it?) Let us consider a fixed voltage. $$I = nAve$$ is the relation between current, number of free electrons in unit volume $$(n)$$, cross-sectional area $$A$$ and electron drift velocity $$v$$. ($$e$$ is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area $$A$$ of the original wire. Neither the $$n$$ quantity, nor $$v$$ is changed since we did not change the material of the wire. Now how the Ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where $$n$$ is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? (Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision, isn't it?) 3 Summary of the question added at conclusion edited May 8 '16 at 22:28 Tony Marshle 1944 bronze badges Let us consider a fixed voltage. I = nAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? (Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision , isn't it?) Let us consider a fixed voltage. I = nAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? Let us consider a fixed voltage. I = nAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? (Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision , isn't it?) 2 added 39 characters in body edited May 8 '16 at 22:13 Tony Marshle 1944 bronze badges Let us consider a fixed voltage. I = nAvnAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? Let us consider a fixed voltage. I = nAv is the relation between current, number of free electrons in unit volume, cross-sectional area and electron drift velocity. We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? Let us consider a fixed voltage. I = nAve is the relation between current, number of free electrons in unit volume (n), cross-sectional area A and electron drift velocity v. (e is the charge of electron) We know how to halve the current by modifying the wire, just by halving the cross sectional area A of the original wire. Neither the n quantity, nor v is changed since we did not change the material of the wire. Now how the ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where n is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance? 1 asked May 8 '16 at 21:19 Tony Marshle 1944 bronze badges