Question

A wire carries a current of 1.A. How many clcc.
trons pass a point in the wire an each second?​

Answer 1

Answer:

• No. of Electrons = 6.25 × 10¹⁸ .

Given:

1. Current (I) = 1 Ampere.
2. Time taken (t) = 1 seconds.

Explanation:

$\rule{300}{1.5}$

From the Charge & Current relation we Know,

$\large{\boxed{\bold{I = \dfrac{Q}{t}}}}$

$\bold{Here}\begin{cases}\text{I represents Current} \\ \text{Q represents Charge} \\ \text{t represents Time taken}\end{cases}$

Substituting the values,

$\large{\boxed{\bold{I = \dfrac{Q}{t}}}}$

$\large{\tt \hookrightarrow 1 \; A = \dfrac{Q}{1 \sec}}$

$\large{\tt \hookrightarrow 1 = \dfrac{Q}{1}}$

$\large{\tt \hookrightarrow Q = 1 \times 1}$

$\large{\boxed{\tt Q = 1 \: Columb}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Relation,

$\large{\boxed{\bold{ne = Q}}}$

$\bold{Here}\begin{cases}\text{n represents No. of Electrons} \\ \text{Q represents Charge} \\ \text{e represents Charge of electron}\end{cases}$

Substituting the values,

$\large{\boxed{\bold{ne = Q}}}$

• Charge on Electron = 1.6 × 10⁻¹⁹ Columb.

$\large{\tt \hookrightarrow n \times 1.6 \times 10^{-19} = 1C}$

As we Got Charge as 1 Columb in the Above equation

$\large{\tt \hookrightarrow n \times 1.6 \times 10^{-19} = 1C}$

$\large{\tt \hookrightarrow n = \dfrac{1}{1.6 \times 10^{-19}}}$

$\large{\tt \hookrightarrow n = \dfrac{1 \times 10^{19}}{1.6}}$

$\large{\tt \hookrightarrow n = \dfrac{10 \times 10^{19}}{16}}$

$\large{\tt \hookrightarrow n = \dfrac{\cancel{10} \times 10^{19}}{\cancel{16}}}$

$\large{\tt \hookrightarrow n = \dfrac{5 \times 10^{19}}{8}}$

$\large{\tt \hookrightarrow n = 0.625 \times 10^{19}}$

$\huge{\boxed{\boxed{\tt n = 6.25 \times 10^{18}}}}$

∴ 6.25 × 10⁻¹⁸ Electrons Flow Through 1 A current.

$\rule{300}{1.5}$