## What is an Electric Current?

Electric charges are responsible for electric current and is defined as the ratio of total charge over a cross sectional area to the time. The SI unit of current is "ampere".

A force is experienced by the electric charge due to the presence of a field around it known as electric field. So, if the charge can move freely in this field then it will constitute an electric current. The free charged particles exist in the ionosphere in the atmosphere. The negative electric charge is termed as electron and the positive electric charge is referred to as nuclei and are not free to move though they are bounded together.

- Several different molecules containing matter are known as bulk matter, for example in grams has contained approximately 1022 molecules because of tight packing electrons.
- Electrons are not in contact with the nuclei.
- Electrons in certain cases still have the acceleration remaining zero in the presence of an electric field.
- In the case of conductors, electrons are free to move in an applied field.
- SI unit of electric current or current is denoted by (A).

## What is Current Density?

The flow of current over an area is known as current density. The area which is considered is normal to the direction of the flow of electric.

### Current and Electron flow

In Current and Electron: The energy is lost from the circuit when the flow of electric charge or charge is in the opposite direction i.e positive towards the negative. Whereas the electrical flow is because of the negative charges that have the direction of the flow from the negative towards the positive. While the flow of electric current is always considered as positive to the negative charges.

State the number of electrons required to produce an electric current of one milliampere?

Solution.

We have, $I=\frac{q}{t}$

From the quantization of charge

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

where n is the number of electrons

$n=\frac{It}{e}=\frac{{10}^{-3}\times 1}{1.6\times {10}^{-19}}=6.25\times {10}^{15}\text{\hspace{0.17em}}\text{electrons}$

## Ohm's Law

From a conductor, if an electric current is passing through then this law states that the current is directly proportional to the potential difference applied and the proportionality sign is replaced by a constant called resistance.

$\begin{array}{c}V\propto I\\ V=IR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(1\right)\end{array}$

Where R denotes the resistance of the conductor.

Also for resistance $R=\frac{V}{I}$

## What is Resistance?

Resistance as the name suggests is the property that resists or restricts the flow of current or electric supply.

Below mentioned are the factor on which the resistance depends:

- The Material
- Length
- Its cross-sectional area
- Its temperature

As per the figure let a conductor have a slab length l and area A of the cross section. If in the next step two slabs that are identical are placed side by side such that the length is now doubled. The electric current will be as similar to that as flowing in any of the individual slabs. Let V be the potential difference and I be the current through the slabs.

$\begin{array}{c}{R}_{c}=\frac{2V}{I}\\ =2R\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{...}\left(2\right)\end{array}$

since $R=\frac{V}{I}$, for any of the slabs.

Thus doubling length will double the resistance i.e

$R\propto l\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{.....}\left(3\right)$

The cross-sectional area will be (A/2) for two identical slabs.

Thus, the current will also in this case be (I/2) while the potential remains the same, i.e., The resistance of each slab is

${R}_{1}=\frac{V}{\left(\frac{I}{2}\right)}=\frac{2V}{I}=2R\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{....}\left(4\right)$

Thus if the area is half the resistance is double

$R\propto \frac{1}{A}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{...}\left(5\right)$

From the equations in 3 and 5 we have

$R\propto \frac{l}{A}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{....}\left(6\right)$

Thus the relation for a given electric conductor

$R=\rho \frac{l}{A}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{......}\left(7\right)$

Here $\rho $ denotes the resistivity

### Drawbacks of Ohm’s law

- Only good conductors follow this law.
- Only in the condition where the physical electrical conditions are constant the law is applicable.
- For the extreme and minimum temperature the law is not applicable.
- For semiconductors, thermistors, vacuum tubes, discharge tubes, law is not applicable.

## What do You Understand by Resistivity?

The material property affected due to resistance it is known as resistivity and is denoted by the ‘’_{ }ρ’’.

The unit of resistivity is Ωm

Resistivity

$\rho =\frac{m}{\left(n{e}^{2}t\right)}$

Here,

'm' is for mass.

'n' for the number density of electrons.

'e' charge of the electron.

'τ' is for the relaxation time of l electrons.

'ρ' is independent of geometric dimensions.

### When Resistors are in Series and Parallel

**For Series **

The electric current in this scenario through two resistors will be same but the potential difference will be calculated as

${V}_{1}=I{R}_{1}$

${V}_{2}=I{R}_{2}$

**For series combination: **

$R={R}_{1}+{R}_{2}$

Question. Determine the electrical voltage drop for each resistor given in the figure for a 10 V supply and the connection of resistors in series.

Solution,

Given,

$\begin{array}{l}{R}_{1}=5\Omega \\ {R}_{2}=3\Omega \\ {R}_{3}=2\Omega \\ V=10\text{\hspace{0.17em}}volt\end{array}$

So, the effective resistance of series combination will be,

${R}_{s}={R}_{1}+{R}_{2}+{R}_{3}=10$

Current in circuit

$I=\frac{V}{{R}_{s}}=\frac{10}{10}=1\text{\hspace{0.17em}}\text{Ampere}$

The voltage drop across R_{1},

${V}_{1}=I{R}_{1}=1\times 5=5V$

The voltage drop across R_{2} ,

${V}_{2}=I{R}_{2}=1\times 3=3V$

The voltage drop across R_{3}

${V}_{3}=I{R}_{3}=1\times 2=2V$

For Parallel Resistors

The potential difference in this scenario will be same through two resistors but the electric current will be calculated as

Resistors in parallel:

$\frac{1}{R}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}$

Question. Determine the current flowing across resistors having value as 3,5 and 2 ohm and also connected in parallel.

Solution.

Given,

R_{1} = 3Ω

R_{2} = 5Ω

R_{3} = 2Ω

Supply voltage V = 15 volt

The effective resistance of parallel combination:

$\begin{array}{l}\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}+\frac{1}{{R}_{3}}\\ \frac{1}{{R}_{p}}=\frac{1}{3}+\frac{1}{5}+\frac{1}{2}\\ {R}_{p}=0.9677\text{\hspace{0.17em}}\Omega \end{array}$

Current through R_{1},

${I}_{1}=\frac{V}{{R}_{1}}=\frac{15}{3}=5\text{\hspace{0.17em}}\text{A}$

Current through R_{2},

${I}_{2}=\frac{V}{{R}_{2}}=\frac{15}{5}=3\text{\hspace{0.17em}}\text{A}$

Current through R_{3},

${I}_{3}=\frac{V}{{R}_{3}}=\frac{15}{2}=7.5\text{\hspace{0.17em}}\text{A}$

Total electricity

$I=\frac{V}{{R}_{P}}=\frac{15}{0.9677}=15.5\text{A}$

### Series and parallel circuits

**Series **

For series the diagram illustrates the combination

**Parallel **

For parallel the diagram illustrates the type of flow.

## Electrical energy

The measurement of voltage between two considered points is known as potential difference. Also, it states when the form of energy i.e. electrical is changed to some other form because of the electric current passed.

The instrument called voltmeter is responsible for measuring the voltage difference between two points.

At A the electricity has a lot of energy but at B most of this energy has been changed into heat and light in the bulb.

$\begin{array}{l}\text{Energy}=\text{Voltage}\times \text{Charge}\\ \text{Energy}=\text{Voltage}\times \text{Current}\times \text{time}\end{array}$

## Common Mistakes

- Making common math errors in the circuit design.
- Misidentifying series and parallel device connected.
- Simplifying circuit incorrectly.

## Formulas

- $\text{Voltage}=\text{Current}\times \text{time}$

- $\begin{array}{l}\text{Energy}=\text{Voltage}\times \text{Charge}\\ \text{Energy}=\text{Voltage}\times \text{Current}\times \text{time}\end{array}$

- $R=\rho \frac{l}{A}\text{\hspace{0.17em}}$

**Context and Application**

In home appliances, for example several electric bulbs connected either in series or parallel connection in a home. Many switches which we used in our daily life all are having electricity whereas the Gadgets which we used all are having electricity or electric current passing through them. It is studied in

- Bachelors and Masters in Science (Physics)
- Bachelors in Technology (Electrical Engineering)

### Want more help with your physics homework?

*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.

# Electric Current Homework Questions from Fellow Students

Browse our recently answered Electric Current homework questions.