Question

A battery of emf $2V$ and internal resistance $0.5\Omega$ is connected across a resistance of $9.5\Omega$. How many electrons cross through a cross-section of the resistance in $1s$?

Answer 1

Step 1: Given Data

Internal resistance $r=0.5\Omega$

Resistance across the resistor $R=9.5\Omega$

Emf of the battery $\epsilon =2V$

Time taken $t=1s$

Step 2: Calculate the Current

According to Ohm's law, we know that

$E=I\left(R+r\right)$

$\Rightarrow I=\frac{E}{R+r}$

Upon substituting the values we get,

$I=\frac{2}{9.5+0.5}$

$=\frac{2V}{10\Omega }$

$=0.2A$

Step 3:

We know that current is defined as the flow of charge $Q$ per unit of time $t$.

$\therefore I=\frac{Q}{t}$

We know that the total charge is the product of the charge of a single electron $e$ and the number of electrons $n$.

$\therefore I=\frac{ne}{t}$

$\Rightarrow n=\frac{It}{e}$

The charge of an electron $e=1.6\times {10}^{-19}C$

Upon substituting the values we get,

$n=\frac{0.2A\times 1s}{1.6\times {10}^{-19}C}$

$=1.25\times {10}^{18}$ electrons

Hence, the number of electrons that cross through a cross-section of the resistance is $1.25\times {10}^{18}$ electrons.