Ohm’s Law for the closed circuit , Relation between emf ( VB ) of an electric cell & voltage across its poles

Ohm’s Law

The emf of an electric cell ( battery – source ) is the total work done inside and outside the cell to transfer electric charges in the electric circuit , If we denote the emf of the battery by ( VB ) , the total current intensity in the circuit by ( I ) , the external resistance by ( R ) and the internal resistance of the cell by ( r ) .

Then  : VB = I R + I r

VB = I ( R + r )

I  =  VB / ( R + r )

This is known as Ohm’s law for a closed circuit where :

Electric current intensity = Total electromotive force / Total resistance of the circuit

 

Ohm's Law

Ohm’s Law

The relation between emf ( VB ) of an electric cell and voltage across its poles ( V ) 

Based on Ohm’s law for a closed circuit :

VB = I R + I r        ,   V = I R

∴  VB = V + I r      ,    ∴  V = VB − I r

From the previous relation , we see that as the external resistance ( R ) increases , The electric current ( I ) passing in the circuit gradually decreases and the potential difference ( V ) between the poles of the cell increases .

The potential difference ( V ) between the poles of the cell becomes equal to the emf of the source ( VB ) , When the value of the current becomes very small , ( I r ) can be ignored .

The emf of an electric cell is greater than the potential difference between the terminal of its external circuit when the circuit is switched on , Because the internal resistance of the electric cell consumes work to pass the current inside the electric cell based on the relation ( VB = V + I r ) and thus ( V < VB ) .

Hence we may define the emf of the cell as :

The emf of the cell ( VB ) is the potential difference across poles of the cell in case of no current in the circuit ( switch is opened ) , or it is the total work done inside and outside the cell to transfer an electric charge of 1 C ( electric charges unit ) in the electric circuit , The emf of the source is measured in Volt .

When the emf of an electric cell = 3 V , The total work done inside and outside the cell to transfer an electric charge of 1 C in the electric circuit = 3 J .

Ohm's Law

Ohm’s Law

In case of one electric cell connected in the circuit :

Where , VB is the reading of the voltmeter on the battery that have an internal resistance of r , the battery is connected on series with a resistance that have a potential difference of V2 and it is connected on series with an ammeter .

If the switch K is closed :

I  =  VB / ( R + r )

I  =  VB − V1 / r

I  =  V2 / R

V2 = I  R   , V1 = VB − I r

If the switch K is opened :

I = 0

V2 = 0 , V1 = VB

In case of two electric cells connected in series in the circuit

Where , V1 is the reading of the voltmeter on the first battery that have an internal resistance of r1 , V2 is the reading of the voltmeter on the second battery that have an internal resistance of r2 ,  the potential difference on the two batteries is V3 .

When the two batteries are connected in the same direction :

I  = [ ( VB )1 + ( VB )2 ] / ( R + r1 + r2 )

V1 = ( VB )1 − I r1

V2 = ( VB )2 − I r2

V3 = V1 + V2

When the two batteries are connected in the opposite directions ,  Where  ( VB )2 < ( VB )1  :

I  = [ ( VB )1 − ( VB )2 ] / ( R + r1 + r2 )

V1 = ( VB )1 − I r1   ( discharging case ) 

V2 =  ( VB )2 + I r2  ( charging case ) 

V3 = V1 V2

Electrical current , Potential difference , Electric resistance and Ohm’s law

Resistances connection ( series & parallel ) , Electric energy and Electric power

Kirchhoff’s first law , Kirchhoff’s second law & how to solve problems of Kirchhoff’s laws

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *