The amount of chemical energy converted to electrical energy per coulomb of charge (C) when passing through a power supply
e.m.f is measured using a voltmeter connected in parallel with the cell
V = IR
The work done per unit charge / coulomb to overcome the internal resistance / resistance inside the battery (when current flows)
v = ε – V = Ir (Ohm’s law)
The resistance of the materials within the battery
Circuit showing the e.m.f and internal resistance of a power supply
A battery of e.m.f 7.3 V and internal resistance r of 0.3 Ω is connected in series with a resistor of resistance 9.5 Ω.Determine:
a) The current in the circuit
b) Lost volts from the battery
If the exam question states 'a battery of negligible internal resistance', this assumes that e.m.f of the battery is equal to its voltage. Internal resistance calculations will not be needed here.If the battery in the circuit diagram includes internal resistance (like that in the worked example), then the e.m.f equations must be used.
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