Electric Current Formula: Definition, Formula & Solved Questions

Jasmine Grover Content Strategy Manager

Content Strategy Manager

Electric Current can be defined as the continuous flow of electrons in an electric circuit. It is referred to as the rate of flow of negative charges of the conductor. When the potential difference is applied across the wire or terminal, there is a movement of the electrons. In simpler terms, electric current is the rate of change of electric charge through a circuit. Electric Current is represented by the symbol I and its SI unit is Ampere. The electric current is related to the electric charge and the time.

Table of Content

What is Electric Current?
Types of Electric Current
Electric Current Formula
Solved Examples of Electric Current Formula
Things to Remember
Previous Year Questions
Sample Questions

Key Terms: Electric Current, Electrons, Electric Flow, Electric Charge, Electric Circuit, Conductor, Potential Difference, Resistance, Amperes, Atoms, Ampere

What is Electric Current?

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There are a large number of free electrons in the electricity-conducting material which move from one atom to the other randomly. The free electrons will move towards the positive terminal of the cell when the potential difference is applied across a wire or terminal. Thus, electric current is basically this continuous flow of electrons. The electric current is measured in Ampere (SI Unit of Current).

The conventional direction of an electric current is the direction in which a positive charge will move, thus, the current flowing in the external circuit is directed away from the positive terminal of the cell to the negative terminal of the cell.

Electric Current

Electric Current

Discover about the Chapter video:

Current Electricity Detailed Video Explanation:

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Types of Electric Current

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Electric current can be classified into two types based on the flow of electric charge:

Alternating Current: The charges flow in both directions in this type of Current.
Direct Current: In a direct current, the charges flow in one direction.

AC and DC Current

AC and DC Current

Read More: Difference between AC and DC

Electric Current Formula

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The current formula is derived from Ohm’s Law. Using Ohm’s Law, the Electric Current Formula is as follows:

I = V/R

Where

I represent Electric Current in Ampere
V is the Potential Difference in Volt
R is the Resistance in Ohm (Ω)

Solved Examples on Electric Current Formula

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Example 1: An electric heater in a house has a resistance of 70 Ω and a potential difference of 220 V. What will be the magnitude of the current flowing through the heater?

Solution: Resistance (R) = 70 Ω

Voltage (V) = 220

The electric current formula as per the Ohm's Law is given as

I = V / R

I = 220 / 70

I = 3.1428 Amperes

Example 2: The electric current flowing in an electric circuit is given as 50Amp and the resistance of the wires is 14Ω. What will be the potential difference as per the current formula?

Solution: Given that

I = 50 A

R = 14Ω

Using the Electric Current Formula as per Ohm's Law

I = V/R

50 = V/14

V = 50 × 14

V = 700 Volts

Thus, the potential difference is 700 V.

Also Read:

Related Links
Ohm's Law and Its Limitations	Electric Charges and Fields	Internal Resistance
Coulomb's Law, Electric Field, Electrostatic Pressure	Unit of Electric Charge	Types of Current
Power of Alternating Current	Unit of Voltage	Electromagnetic Induction & Alternating Currents

Things to Remember

Electric current is defined as the rate of change of electric charge through a circuit.
Electric Current relates to the voltage and resistance of the circuit. It is a scalar quantity.
The electric current is represented by the symbol I and its SI unit is Ampere.
According to Ohm’s Law, the electric current formula is: I = V/R
An electric wire carries a current of 1 ampere when a charge flows through it at the rate of one coulomb per second.

Previous Year Questions

When the switch S , in the circuit shown, is closed, then the value of current ii will be…. [ JEE Main 2019]
The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor… [JEE Main 2019]
In the given circuit the cells have zero internal resistance…. [JEE Main 2019]
The currents I2,I3 and I6,respectively, are….[JEE Main 2019]
A d.c. main supply of e.m.f. 220V is connected across a storage battery of e.m.f. 200V through a resistance of 11 Ω…... [JEE Main 2014]
The percentage change in its electrical resistance if its volume remains unchanged is…. [JEE Main 2019]
three resistors 3Ω,9Ω and 9Ω and a capacitor….. [JEE Main 2013]
If now we hav.e to change the null point at 9th wire, what should we do?…
The electrical permittivity and magnetic permeability of free space are…
In the given circuit, an ideal voltmeter connected across the 10Ω resistance reads 2V . The internal resistance rr, of each cell is….[JEE Main 2019]
In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance…. [JEE Main 2019]
Correct set up to verify Ohm’s law is….[JEE Main 2013]
The resistance between any two terminals is when connected in a triangle is…. [ NEET 1993 ]
potential drop through 4Ω resistor is… [ NEET 1993 ]
The potential difference per unit length of the wire will be… [ NEET 1999 ]
Value of R for which the power delivered in it is maximum is given by... [ NEET 1992 ]

Sample Questions

Ques. Calculate the electric current passing through the circuit in which the voltage and resistance be 25V and 5 ω respectively? (3 Marks)

Ans. V = 25 V

R = 5ω

Now, here we will apply ohm’s law.

The equation for calculating the electric current using Ohm’s law is given as

I = V/R

Putting the known values, we get

I = 25/5V

I = 5 A

Therefore, the value of electric current passing through the circuit in which the voltage and resistance be 25V and 5 ω respectively is 5 A.

Ques. A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 39 Ω. Find the value of the current?. (3 Marks)

Ans. According to Ohm’s Law,

I = V/R

Putting the given values, we get

The Value of current, I = 200V - 5V /39 Ω

Therefore, the value of current = 5 A.

Ques. The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery? (3 Marks)

Ans. In the given question,

The emf of the battery is given as E = 12V

The internal resistance of the battery is R= 0.4 Ω

The amount of maximum current drawn from the battery is given as I

According to ohm's law,

E = IR

Rearranging we get,

I = E/R

I = 12V/ 0.4 Ω

I = 30 A,

Therefore, maximum current drawn from the battery is 30 A.

Ques. What are the types of electric current? (3 Marks)

Ans. There are two types of electric current namely:

Direct Current (DC)
Alternating Current (AC)

With direct current, electrons move in one direction. Batteries produce direct current. In alternating current, electrons move in both directions.

Ques. A battery of Emf 10V and internal resistance 3 Ω is connected to a resistor. If the current in a circuit is 0.5A. What is the resistance of a resistor? Calculate what will be the terminal voltage of a battery when the circuit is closed. (5 Marks)

Ans. The Emf of the battery is (E) = 10V

Internal Resistance of the battery (R) = 3 Ω

The current in the circuit is (I) = 0.5A

Consider the resistance of a resistor by R

So, by using ohm's law,

I = E/ R+r

By Rearranging we get,

R+r = E/ I

R+r = 10/0.5

R + r = 20 Ω

R = 20 - 3 Ω = 17 Ω

Terminal voltage of the resistor is V

As per Ohm's Law,

V = IR

V = 0.5×17

V = 8.5 V

Thus, the resistance of the resistor is 17Ω and the terminal voltage of the battery when the circuit is closed is 8.5 V.

Ques. 2-ampere current is flowing through a conductor having 2-ohm resistance. Calculate the potential difference across a conductor? (3 Marks)

Ans. Here,

I = 2 ampere

R = 2 ohm

Now, V = I R

V = 2 x 2

V = 4 volts

Thus, The potential difference across the conductor is 4 Volts.

Ques. The voltage & resistance of a circuit is 20V and 2Ω respectively. Calculate what will be the value of the current through the circuit. (3 Marks)

Ans. Here,

R = 2 Ω

V = 20 V

We know that according to Ohm’s Law

V = I R

or I = V/R

I = 20/2 = 10 Amperes

Thus, the value of the current through the circuit is 10 Amperes.

Ques. The voltage and resistance of a circuit are given as 10V and 4Ω respectively. Calculate the current through the circuit. (3 Marks)

Ans. The given values are,

V = 10V

R = 4Ω

The equation for electric current using Ohm’s law is,

I = V/R

I = 10/4

= 2.5 Ampere

Thus, the value of the current through the circuit is 2.5 Ampere.

Ques. What is Electrical Resistance? (1 Mark)

Ans. Resistance from power otherwise called electrical opposition is a power that neutralizes the progression of flow.

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CBSE CLASS XII Related Questions

1.
The resistance of a wire at 25°C is 10.0 $ \Omega $. When heated to 125°C, its resistance becomes 10.5 $ \Omega $. Find (i) the temperature coefficient of resistance of the wire, and (ii) the resistance of the wire at 425°C.
View Solution
2.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 $\Omega$), (2 V, 4 $\Omega$) and (6 V, 2 $\Omega$) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.
View Solution
3.
The electric field at a point in a region is given by $ \vec{E} = \alpha \frac{\hat{r}}{r^3} $, where $ \alpha $ is a constant and $ r $ is the distance of the point from the origin. The magnitude of potential of the point is:
- $ \frac{\alpha}{r} $
- $ \frac{\alpha r^2}{2} $
- $ \frac{\alpha}{2r^2} $
- $ -\frac{\alpha}{r} $
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4.
A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle $i$ such that it is refracted out grazing the face AD. Find the value of angle $i$.
$A rectangular glass slab ABCD (refractive index 1.5)$
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5.
(a) Consider the so-called ‘D-T reaction’ (Deuterium-Tritium reaction).
In a thermonuclear fusion reactor, the following nuclear reaction occurs: \[ \ ^{2}_1 \text{H} + \ ^{3}_1 \text{H} \longrightarrow \ ^{4}_2 \text{He} + \ ^{1}_0 \text{n} + Q \] Find the amount of energy released in the reaction.
% Given data Given:
$ m\left(^{2}_1 \text{H}\right) = 2.014102 \, \text{u} $
$ m\left(^{3}_1 \text{H}\right) = 3.016049 \, \text{u} $
$ m\left(^{4}_2 \text{He}\right) = 4.002603 \, \text{u} $
$ m\left(^{1}_0 \text{n}\right) = 1.008665 \, \text{u} $
$ 1 \, \text{u} = 931 \, \text{MeV}/c^2 $
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6.
A system of two conductors is placed in air and they have net charge of $ +80 \, \mu C $ and $ -80 \, \mu C $ which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
(3) If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.
View Solution