To transfer the maximum power from a generator into a load the impedance of the load and the internal impedance of the generator – and any intervening transmission line – must be equal.
Figure 3.1 illustrates the simplest case of a generator of internal impedance Zs equal to 5 ohms and producing an e.m.f. of 20 volts.
When loads of varying impedance, Z1, are connected the output voltage, V (p.d.) and the power in the load, P1, varies as follows:
Z1 = 5I = 20/10 =2A V = 10 V Z1 = 3I = 2.5A V = 7.5V Z1 = 8.33I = 1.5A V = 12.5V P1 = V 2/Z1 = 100/5 = 20 W
P1 = V 2/Z1 = 56.25/3 = 18.75 W
P1 = V 2/Z1 = 156.25/8.33 = 18.75 W
When dealing with alternating current and when transmission lines, particularly at radio frequencies, are interposed between the source and the load, other factors than the power transfer efficiency must also be considered.
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