Answer :

We know an cell has some internal resistance of its own due to which potential difference across cell is not as marked on it but a bit less, the marked voltage of the cell is called emf of the cell which is the potential difference across the cell when cell is not connected to external resistance or no current is being drawn from it.

Now here we have a series combination of three cells, let emf of each cell be E and internal resistance of each be r , suppose the combination is connected to an external resistance R , since cells are in series their emf’s will be added so net emf of the cell will be E_{net} = 3E

Internal resistances of the cells are also in series so net internal resistance of the combination is

r_{net} = 3r

**The circuit diagram is as Shown**

Now we know according to ohm’s law, the potential difference across a resistor is equal to product of current flowing through it and its Resistance Mathematically,

V = IR

Where, V is the potential difference across the ends of register, I is the current flowing through it and R is the resistance

Now potential difference across the net internal resistance of the battery or combination of cells is

V_{1} = Ir_{net}

Where I is the current flowing through the circuit and r_{net} is the net internal resistance

In any resistance current flow from higher potential to lower potential so polarity of voltages of cell **is as shown**

If we let the net potential difference across battery be V so

V = E_{net} – V_{1}

Or

V = E_{net} – Ir_{net}

Now when current drawn from cell is 0A

i.e. I = 0 A

V = E_{net}

Or we can say when no current is drawn from cell, its voltage is equal to emf of the cell

From the graph we can see

When current in the circuit , I = 0 A

Voltage is V = 6 V

i.e. net emf of the cells is

E_{net} = 6 V

Or we can say

3E = 6 V

i.e. E = 6/3 V = 2 V

so emf of each cell is 2 V

now also from the graph

when voltage across combination of cell V = 0 V

current the circuit, I = 1 A

i.e. 0 V = E_{net} - Ir_{net}

or E_{net} = Ir_{net}

6 V = 1 A × r_{net}

So net internal resistance is

r_{net} = 6 Ω

or we can say

3r = 6 Ω

i.e. r = 6/3 = 2 Ω

so internal resistance of each cell is 2 Ω

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