Question

A 9.0 V battery is connected to a bulb whose resistance is 1.6 Ω. How many electrons leave the battery per minute?

Answer 1

Concepts and reason

The required concept to solve the given problem is Ohm’s law.

First find the total charge in the battery by using number of electrons and charge of the electron and then calculate the number of electrons passing through the battery in a minute by using total charge in the battery.

Fundamentals

Use Ohms law to find the current passing through the resistor.

$\begin{array}{c}\\V = IR\\\\I = \frac{V}{R}\\\end{array}$

Here, V is the potential, R is the resistor and I is the current passing through the resistor.

Total charge in the battery if the number of electrons is n is,

$Q = ne$

Here, e is the charge of the single electron.

When current I passing through the resistor R then the voltage developed in the circuit is,

$V = IR$

Rearrange the above equation for current.

$I = \frac{V}{R}$

Substitute 9V for V and $1.6\Omega$ for R in the above equation.

$\begin{array}{c}\\I = \frac{{9V}}{{1.6\Omega }}\\\\ = 5.625A\\\end{array}$

The rate of change of the charge flow is called current.

$I = \frac{{dQ}}{{dt}}$

Rearrange the above equation for Q.

$\begin{array}{c}\\\int {dQ} = I\int {dt} \\\\Q = I\,t\\\end{array}$

Substitute $5.625A$for I and 1min for t in the equation above equation to get charge flow per minute.

$\begin{array}{c}\\Q = \left( {5.625A} \right)\left( {1min} \right)\\\\ = \left( {5.625A} \right)\left( {1min\left( {\frac{{60sec}}{{1min}}} \right)} \right)\\\\ = 337.5C\\\end{array}$

The charge of the electron is,

$e = 1.6 \times {10^{ - 19}}C$

The number of electrons in the charge $337.5C$is,

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

Here, Q is the total charge and e is the charge of the single electron.

Substitute $337.5C$for Q and $1.6 \times {10^{ - 19}}C$for $e$in the above equation.

$\begin{array}{c}\\n = \frac{{337.5C}}{{1.6 \times {{10}^{ - 19}}C}}\\\\ = 210.93 \times {10^{19}}\\\\ = 2.1 \times {10^{21}}\\\end{array}$

Therefore, the number of electrons leave the battery per minute is$2.1 \times {10^{21}}$.

Ans:

The number of electrons leave the battery per minute is$2.1 \times {10^{21}}$.