Question

given that the strength of circuit is 2 a. find the no of electrons passing through the cross section in 1 sec

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

We have a relation between charge and current as

Charge=Current×time i.e Q=i×t.

Now we have current =10amp & time=1sec from given data.

Therefore charge Q=10×1=10 coulombs.

Again we have a relation between charge and no of electrons

I.e no of electrons= total charge/1.6×10^-19.

So, no of electrons=( 10C/1.6×10^-19 )=6.25×10^19 electrons.

Therefor, 6.25×10^19 electrons will flow through a copper wire carrying 10amps current for 1sec.

Explanation: