**Electric Current:**The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor.

By convention, electric current flows in opposite direction to the movement of electrons.

**Electric Circuit:**Electric circuit is a continuous and closed path of electric current.

**Expression of Electric Current:**Electric current is denoted by letter ‘I’. Electric current is expressed by the rate of flow of electric charges. Rate of flow means the amount of charge flowing through a particular area in unit time.

If a net electric charge (Q) flows through a cross section of conductor in time t, then;

Where,I is electric current,Q is net charge and t is time in second.

SI unit of electric charge is coulomb (C).

One coulomb is nearly equal to 6 x 1018 electrons. SI unit of electric current is ampere (A). Ampere is the flow of electric charges through a surface at the rate of one coulomb per second. This means if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Therefore; 1 A = 1 C/1 s

**Small quantity of Electric Current:**Small quantity of electric current is expressed in milliampere and microampere. Milliampere is written as mA and microampere as μA

1mA (milliampere)= 10−3 A

1μA(microampere)=10−6 A

**Ammeter:**An apparatus to measure electric current in a circuit.

**Solution:**Given, electric current (I) = 5 A

Time (t) = 5 minute = 5 X 60 = 300 s

Electric charge (Q) =?

We know; I = Q/t

Or, Q = I x t

Or, Q = 5 A x 300 s = 1500 C

**Solution:**Given, electric current (I) = 2 A

Time (t) = 1 hour = 1 x 60 x 60 s = 3600 s

Electric charge (Q) =?

We know that Q = I x t

Therefore, Q = 2 A x 3600 s = 7200 C

**Solution:**Given, electric charge (Q) = 6000 C

Electric current (I) = 10 A

Time (t) = ?

We know; I = Q/t

Or, t = Q/I

Or, t = 6000 C ÷ 10 A = 600 s

Or, t = 10 min

**Solution:**Given, electric charge (Q) = 900 C

Time (t) = Half an hour = 30 m = 30 x 60 = 1800 s

Electric current (I) =?

We know; I = Q/t

Or, I = 900 C ÷ 1800 s = 0.5 A

**Solution:**Given, electric charge (Q) = 1500 C

Time (t) = 5m = 5 x 60 = 300 s

Electric current (I) =?

We know; I = Q/t

Or, I = 1500 C ÷ 300 s = 5 A

# Electric Potential and Potential Difference

**Electric Potential:**The amount of electric potential energy at a point is called electric potential.

**Electric Potential difference:**The difference in the amount of electric potential energy between two points in an electric circuit is called ELECTRIC POTENTIAL DIFFERENCE.

Electric potential difference is known as voltage, which is equal to the work done per unit charge to move the charge between two points against static electric field.

Example 4: Calculate the potential difference between two points, if 1500 J of work is done to carry a charge of 50C from one point to other?Therefore;

Voltage or electric potential difference is denoted by ‘V’. Therefore;

Where, W = work done and Q = Charge

#### SI unit of electric potential difference (Voltage):

SI unit of electric potential difference is volt and denoted by ‘V’. This is named in honour of Italian Physicist Alessandro Volta.

Since joule is the unit of work and coulomb is the unit of charge; 1 volt of electric potential difference is equal to the 1 joule of work to be done to move a charge of 1 coulomb from one point to another in an electric circuit. Therefore,

Voltmeter: An apparatus to measure the potential difference or electric potential difference between two points in an electric circuit.

**Solution:**Given, potential difference (V) = 15 V

Charge (Q) = 5 C

Work done (W) =?

We know that; V = W/Q

Or, W = V x Q

Or, W = 15 V x 5 C = 75 J

**Solution:**Given, charge = 3 C

Potential difference between two points = 10V

Work done (W) =?

We know that; V = W/Q

Or, W = V x Q

Or, W = 10 V x 3 C = 30 J

**Solution:**Given, work done (W) = 100J

Charge (Q) = 10C

Potential difference (V) =?

We know that; V = W/Q

Or, V = 100 J ÷ 10 C = 10 V

**Solution:**Given, work done (W) = 1500J

Charge (Q) = 50C

Potential difference (V) =?

We know that; V = W/Q

Or, V = 1500 J ÷ 30 C = 50 V

**Solution:**Given, potential difference (V) = 100V

Work done (W) = 5000 J

Charge (Q) =?

We know that; V = W/Q

Or, Q = W/V

Or, Q = 5000 J ÷ 100 V = 50 C

**Solution:**Given, potential difference (V) = 220V

Work done (W) = 1760 J

Charge (Q) =?

We know that; V = W/Q

Or, Q = W/V

Or, Q = 1760 J ÷ 220 V = 8 C

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