In the electrical circuits we have looked at so far, we have regarded the voltage source supplying the circuit as being a perfect voltage source. i.e.:
In practice this is not the case as:
This is illustrated in the animation below.
This variation of supply voltage is potentially a difficult thing to explain, especially as the actual
reasons for the variation would depend on the type of power supply being used.
For example:
Fortunately we can avoid these details and regardless of the actual nature of the power supply, we can represent it with a power supply model which consists entirely of simple electrical components. One way of doing this is by representing the power supply as a perfect voltage source, (an e.m.f.) in series with an internal resistance. When this power supply model is applied to an external circuit, then the circuit current also flows through the internal resistance. This produces an internal voltage drop inside the power supply, which therefore reduces the voltage across the power supply terminals. The power dissipated by the internal resistance, represents the heat generated in the power supply. This is illustrated in the animation below.
The terminal voltage (V) is equal to the e.m.f. voltage (E), minus the internal voltage drop (Ir).
(using ohms law: internal voltage drop = current (I) x internal resistance (r)).
To model any real power supply, we just have to determine the correct values of E and r to use.
When the power supply is not connected to a circuit, there will be no current flowing, therefore:
The internal resistance can be determined, by connecting a circuit of known resistance and measuring the current that flows.
The power supplied by the e.m.f., is given by P = EI and the power dissipated in the power supply, is given by P = I^{2}r.
The energy provided by the e.m.f., is given by W = EIt and the energy dissipated in the power supply, is given by W =I^{2}rt.