Current Electricity 

Current electricity can be referred to as the flow of electric charge (electrons) through conductors. Electric current (I) is defined as the flow of electric charge. Current electricity is basic for understanding circuits, resistance, power, and other electrical phenomena.

1.0Basics of Current Electricity

Conventional Current

Conventional current is the hypothetical flow of positive charge from the positive to negative terminal of a circuit. Factually, it is electrons, which possess the negative charge and produce electricity that flows from the negative terminal to the positive terminal, against the flow of conventional current. 

Definition of Electric current

Electric current (I) is referred to as the rate of flow of any charge Q through a conductor. The device used for producing electric current (I) is called an Electric Generator. The electric current formula is written as: 

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

Here: 

• I = electric current in amperes
• Q = charge in coulombs
• t = time in seconds. 
• The SI unit of electric current is ampere. 
• Electric Current is a scalar quantity. 

According to Ohm’s law, the formula for electric current is $I=\frac{V}{R}$ where I is the electric current and V and R are the potential difference and resistance of the current-carrying conductor, respectively. 

Related Video:

2.0Types of Electric Current 

The electric currents are of two types based on the flow of electric charge inside any conductor. Electric currents are of these two types: 

• Direct Current (DC): DC is the flow of electric charge in one constant direction. It is produced by sources like batteries and solar cells. The voltage remains steady over time, making it suitable for electronic devices like laptops and phones.

• Alternating Current (AC): AC is the flow of electric charge periodically reversing direction. This is the type of current utilised in homes and industries that is provided by power grids. It can easily send power over large distances because the voltage changes sinusoidally.

3.0Drift Velocity 

Drift Velocity is the average velocity of free charge carriers, usually electrons, in a conductor under the influence of an applied electric field. It is much smaller than the random thermal velocity but results in the net flow of current. The formula for drift velocity v_d is: 

$I=nAe{v}_d$

Where: 

• I represent the electric current, 
• n is the no. of free charge carriers per unit volume V, 
• A is the cross-sectional area of the conductor, 
• e is the charge of an electron, 
• v_d is the drift velocity. 

The relationship between electric current and drift velocity is given by the above equation. The equation shows that the current is directly proportional to the drift velocity. 

4.0Effects of Electric Current

Magnetic Effect of Electric Current 

• An electric current flowing within a conductor generates a magnetic field outside the conductor. This was the basis of electromagnets, which are used in items like motors and transformers. The direction of the magnetic field (B) around a current-carrying conductor can be determined by the following rule: 
• Right-Hand Rule: In this, the direction of the magnetic field (B) around a current-carrying conductor can be determined using the right-hand thumb rule. In case the thumb of the right hand is in the direction of the current, then curled fingers will always show the direction of the magnetic field.

Chemical Effect of Electric Current

• Any electric current passing through a conducting solution, known as an electrolyte, causes a chemical reaction at the electrodes. This process is used in electrolysis to decompose compounds, such as the decomposition of water into hydrogen and oxygen.
• Some examples of this effect in practice include electroplating, electrorefining, and water electrolysis.

Heating Effect of electric current

• When an electric current (I) passes through a conductor, it generates heat that can be utilised in such appliances as electric heaters, toasters, and electric bulbs. Heating occurs as a result of resistance within the conductor; this converts electrical energy into heat energy. The heat (H) generated by an appliance can be calculated by the product of the power dissipated (P) by the appliance in a certain amount of time(t). 

H=Pt

We know that $P=VI=I^{2}R=V^2/R$

(where, V = potential difference (Voltage) and I = Current)

Hence,

$H=VIT=I^{2}Rt=\frac{V^{2}t}{R}$

5.0Solved Examples

Problem 1: A copper wire of length 2 m and cross-sectional area A=1×10⁻⁶ m² carries a current of 3 A. If the no. of free electrons per unit volume in copper is n = 8.5×10²⁸ m⁻³, find the drift velocity of electrons.

Given: $I=3A,n=8.5\times 10^{28}{m}^{-3},A=1\times 10^{-6}{m}^2,e=1.6\times 10^{-19}C$ 

Solution: By using the formula for drift velocity:

$I=nAe{v}_d$

$3=8.5\times 10^{28}1.6\times 10^{-19}1\times 10^{-6}\times v_d$

$v_d=\frac{3}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 1\times 10^{-6}}$

$v_d=\frac{3}{\left(8.5\right)(1.6)(10^4)}$

$v_d=\frac{3}{1.36\times 10^5}$

$v_d=2.21\times 10^{-5}m/s$

Problem 2: A battery with an EMF of 10 V and internal resistance of 1 Ω is connected to two resistors. The resistors have resistances of 3 Ω and 6 Ω, respectively. The resistors are connected in parallel, and this combination is connected in series with the internal resistance of the battery. Find the total current (I) supplied by the battery.

Given: Emf of the battery = 10V, internal resistance, r = 1 Ω, External resistors R₁ = 3Ω and R₂ = 6Ω. 

Solution: Total resistance of the parallel combination: 

$\frac{1}{R_{parallel}}=\frac{1}{R1}+\frac{1}{R2}=$

$\frac{1}{R_{parallel}}=\frac{1}{3}+\frac{1}{6}$

$\frac{1}{R_{parallel}}=\frac{3}{6}=\frac{1}{2}$

$R_{parallel}=2\Omega$

Total resistance in the circuit:

$R_{total}=R_{parallel}+r=2+1=3\Omega$

The current supplied by the battery = I 

Using Ohm’s law: 

$I=\frac{\epsilon }{R_{total}}$

$I=\frac{10}{3}=3.33A.$

Problem 3: A resistor of resistor R=10 is connected to a battery. The battery delivers a total charge of Q=1000 C for 30 min. Calculate the current flowing through the resistor, also calculate the power dissipated by the resistor.

Given: Electric charge Q = 1000C, time (t) = 30 min = 1800s

Solution: The current flowing through the resistor: I 

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

$I=\frac{1000}{1800}$

I=0.56A

The power dissipated by the resistor: P 

$P=I^{2}R$

$P=0.56^2\times 10$

P =3.136W

Also Read