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Electromotive force  (e.m.f.) 

Circuit below shows set up of an electric source of e.m.f. = E volts, driving electric charges round the closed the circuit. The e.m.f. drives the electic chrges through the internal resistance of source as well as the external resistance of the circuit. Hence if the e.m.f. of the cell is 2 V, the total energy dissipated (given out) is 2 joules when one coulomb of charge flows through the closed circuit. Since energy = charge (Q) X potential difference (V). The energy given out will be in the form of heat and light at the bulb with some heat generated within the cell, and along the connecting wires. If Ohm's law is applied for the whole circuit, we have E = I (r + R) i.e. current X total resistance in the whole circuit Where E is the e.m.f. of the source in volts I the current flowing through the closed circuit in the lamp r the internal resistance of the cell in ohm R the total external resistance in ohm 





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