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