Question

A current of 7.5 A is maintained in wire of 45 s. In this time a. how much charge and (b) how many electrons flow through the wire?

Answer 1

Explanation:

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Answer 2

Given: I= 7.5 A and t= 45 s

(a) To find the charge, use the expression of current in terms of charge.

(b)To calculate the number of electrons, use the expression of charge in terms of electrons.

Explanation:

(a)

The relation between the current and the charge is as follows as;

$I=\frac{q}{t}$

Here, I is the current, q is the charge and t is the time.

It is given in the problem that I= 7.5 A and t= 45 s.

Rearrange the above expression for charge to get the value of amount of charge.

q= It

Put I= 7.5 A and t= 45 s.

q= (7.5)(45)

q= 337.5 C

Therefore, the value of charge is 337.5 C.

(b)

The expression for the charge in terms of number of charges and electrons is as follows;

q= ne

Here, n is the number of charges and e is the electron.

Calculate number of electrons by rearranging the above expression for n.

$n=\frac{q}{e}$

Put q= 337.5 C and $e=1.6\times10^{-27}C$.

$n=\frac{337.5}{1.6\times10^{-27}}$

$n=210.4\times10^{27}$

Therefore, the number of electrons flows through the wire is $n=210.4\times10^{27}$.