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Electromotive Force (EMF) and Voltage
No current flows in a copper wire by itself, just as water in a horizontal tube does not flow. If one
end of the tube is connected to a tank with water such that there is a pressure difference between the
two ends of the horizontal tube (A → B), water flows out of the other end (B) at a steady rate. The
rate at which water flows out depends on the pressure difference (h), for a given tube. If the flow
rate (current) is to be kept constant, the water flowing out for instance has to be put back into the jar
to maintain the pressure head, h. This requires work to be done by an external agency. The above
analogy brings out several features of electrical current flow. An electric current flows across a
conductor only if there is an electric potential difference between its two ends. To maintain a steady
current flow, one needs an agency, which does work on the charges. This agency is called the
electromotive force or emf.
In the case of water flow, the agency is the pump 'P' (producing a 'hydromotive force') which
does work at a steady rate in putting the water back into the tank. Just as for a given tube, the
current of fluid flowing out depends on its viscosity (Flow Resistance), the electrical current
flowing for a given potential difference depends on the electrical resistance of the conductor.
If the pump has no internal losses (pressure drops) the hmf is equal to the hydraulic pd (hpd) across
AB.
Our hydraulic Ohm's Law says
hpdAB = flowAB x FlowResistanceAB
(Electrical: V = I x R)
If the pump has internal losses then it has 'internal (flow) resistance' and the hpd will less than the
hmf.
Then InternalFlowResistancePUMP = ( hmfPUMP – hpdAB ) / flowAB
(Electrical: InternalResistanceBAT = (emf – pd) / currentBAT)
'A', 'B' and 'P' corespond to the similar points in the hydraulic system. The correspondence
between the hydraulic and electrical systens would be better if, in the first diagram, the upper
tank were removed and the pump connected directly to 'A'. Then there is no question that
'Flow 1' equals 'Flow 2'.
The above circuit diagram shows the flow of charges. A steady electric current I flows through the
resistance R, from A to B. That is, positive charges flow from higher potential (A) to lower potential
(B). The potential drop from A and B is V. The source P of emf does work on these charges, as they
come through at B because it has to take the positive charges from lower to higher potential. The
charge is transferred from one end of the source of emf to the other and 'qV' work is done on the
charge. The source of emf by doing work on the electric charges, maintains a potential
difference V between its terminals.
The key concept is that the battery, generator or any other source of
emf in an electrical circuit is a "Charge Pump"