E.m.f. And Internal Resistance

Resistance comes from electrons colliding with atoms and losing energy to other forms.
Battery - chemical energy makes electrons move. As they move, they collided with atoms inside, batteries must have resistance = internal resistance. Makes batteries and cells warm up.
Load resistance - total resistance of all components.
Electromotive force - about of electrical energy battery produces for each coulomb of charge. Emf (V) is not force.
P.d across load resistance (R) = energy transferred when 1 coulomb of charge flows through load resistance. This p.d = terminal p.d (V)
No internal resistance of terminal p.d = e.m.f. Real power supplies always lose some energy overcoming internal resistance.
Energy wasted per coulomb overcoming internal resistance = lost volts (v)
Conservation of energy:-

Energy per coulomb supplied by source = energy per coulomb transferred in load resistance + energy per coulomb wasted in internal resistance.

ε = V + v
V = ε - v
ε = I(R+r)
V = ε - Ir

E.m.f (ε)
Terminal p.d (V)
Lost volts (v)
Current (I)
Load resistance (R)
Internal resistance (r)

For cells in series - calculate total e.m.f by adding their individual e.m.f.s:

ε total = ε1 + ε2 + ε3 + ...

Requires all cells to be connected in same direction - one is connected in opposite direction - subtract its e.m.f.

For identical cells in parallel - total e.m.d of combination of cells is the same size:

ε total = ε1 = ε2 = ε3 ...

Vary circuit’s current by changing load resistance using variable resistor. Measure p.d for different values of current. Include a switch to turn off when needed to reduce heating effect in wires on resistance of circuit.
Record V and I in table and plot graph of V against I.
To find e.m.f and internal resistance, start with the equation: V = ε - Ir
Rearrange = V = - rI + ε
Y = mx + c

Intercept = ε

Gradient = -r

Choosing values for load resistance - balancing act. Low load resistance - large current, reduces percentage uncertainty in ammeter reading. Large currents cause significant wire heating - invalidates results.
For experiment - assume voltmeter - high resistance + ammeter resistance = low.
Voltmeters - V. High internal resistance - current = low, can be assumed to be negligible (0). Including voltmeter doesn’t affect current through variable resistor.
Ammeter - low (negligible) resistance, therefore voltage is negligible. Ammeter in circuit doesn’t affect p.d across variable resistor.
Easier way to measure e.m.f of power source - connect voltmeter across terminals. Current though voltmeter = negligible so any difference between measurements + e.m.f will be small therefore difference = insignificant.